1.1 Significant Figures
Definition
The digits in a number which show reliability in measurement
are known as significant figures. They reliable digits known with certainty in
a given number.
OR
The certain digits of a measured quantity plus one uncertain
last digit are called Significant Figures.
Rules for determining the significant
figures
1. All non
zero digits are significant. e.g.
(i) 72 g has two significant figures.
(ii) 124 m has three significant figures.
2. Zeroes lying in between non-zero digits are
significant. e.g.
(i)3002 g has four significant figures.
(ii)16.051 kg has five significant
figures.
3. Final
zeros to the right of the decimal point are significant. e.g.
(i) 3.00 m has three significant figures.
(ii)
6.900 km has four significant
figures.
4. Zeroes
locating the decimal point in number less than one are not significant. e.g.
(i) 0.0004 mm has 1 significant figure.
(ii)
0.00021 mg has 2 significant
figures.
5. Zeros locating
the decimal point in number greater than one are not significant. e.g.
(i) 720 kg has 2 significant figures.
(ii) 472000 miles has 3 significant figures.
1.2 Rounding Off Data
Definition
In chemical calculations, data contains too many digits
and for convenience, we often round off the data into proper number of digits
by dropping last digit (d). Generally, number are rounded to the nearest ten,
hundred, thousand, million and so on. It is a method of reducing the numbers
from the given numerical value.
To reduce a number upto desired significant figures and
adjust the last reported digit is known as rounding off data. It is the
procedure or operation of dropping off last non-significant digits of a data to
reduce a number to the required significant digits thereby adjusting the last
digit reported
The right most digit is generally considered as uncertain,
therefore, we conveniently drop it and round off the number into small numbers
to ensure the maximum removal of errors from the final answer.
Rules for Rounding off Data
1. If last dropping digit is greater than five, then the last remaining digit to be retained is increased by one unit
e.g.
5.768 is rounded off to 5.77 to 3 significant digits.
2. If last dropping digit is less than five, then the last remaining digit will remain unchanged
e.g.
5.734 is rounded off to 5.73 to 3 significant digits (no change in preceding number).
3. If last dropping digit is exactly five, then the last remaining digit is increased by one unit if it is odd and remain unchanged if it is even
e.g.
7.865 is rounded off to 7.86 to 3 significant digits [L.R.D. = even]
8.775 is rounded off to 8.78 to 3 significant digits [L.R.D. = odd]
1.3 Exponential Notation
Definition
While dealing scientific work, it is difficult to use very
small or very large numbers in calculation and it may give wrong results due to
the mistakes in writing too many digits. These larger or small number are much
more easily expressed as multiple of 10 in exponential notation.
For example; the mass of electrons is 0.000, 000, 000, 000, 000, 000, 000, 000, 000, 911 g which is appropriately expressed as 9.11 x 10−28 g.
602,000,000,000,000,000,000,000 representing Avogadro’s
Number, is more conveniently written in exponential notation as 6.02 x 1023
The shorthand mathematical expression
of a very large or a very small number by means of exponents is called
Exponential or Scientific Notation. It is written as Xn where x is
multiplied itself by n time.
Significance
exponential notation is time saving, space saving and helps
in simplification of calculation. It makes calculation simple and easy. It is
frequently used in different fields like physics, engineering, geology,
astronomy etc.
General Representation
In exponential notation, the
numbers are expressed as the product of two numbers; N x 10±x ;
Where,
N = Co-efficient No. or digit term
ranges 1 to 9.999
‘x’ = Positive or negative integer
called Exponent Power raised to base 10
Standard Scientific Notation
Standard Scientific Notation is one in which decimal point
is after one digit of co-efficient number.
For example; 4.56 x 106 is a standard
scientific notation while 45.6 x 105 is not a standard scientific
notation”.
Inter Converting Standard and Scientific Notation
Rule # 1
If we want to convert a numerical value from standard
notation to exponential notation, decimal point should move to the left if the
value is greater than 10 and move to right if the value if the value is in
between 0 to 1. e.g.
4600,000 change to 4.6 x106 (decimal point
moves six places to the left)
0.00038 change to 3.8 x 10−4 (decimal point
moves four places to the right)
Rule # 2
If we need to change an exponential value into standard notation,
the decimal point should move to the right for the positive exponent and move
to the left for negative exponent. e.g.
7.53 x104 can change into 75300 (decimal point
moves four places to the right)
48.7 x 10−5 can change into 0.000487 (decimal
point moves five places to the left)
Rules for determination
of numerical value of exponents
The numerical value of exponents is
determined by following Rules:
Rule I
If the
number to be expressed is greater than 1, the exponent is positive integer (x
>1).
Exponent is positive, when decimal
point is shifted towards left.
The exponent is numerically equal
to the number of places the decimal point has been moved.
e.g. 1 becomes 10°, 10 becomes 101.
Rule II
If the
number to be expressed is less than 1, the exponent is negative integer (x
<1).
Exponent is negative, when decimal
point is shifted to right.
The exponent is equal numerically
to the number of places the decimal point has been moved.
e.g. 0.1 becomes 10–1,
0.01 becomes 10–2
Applications of Exponential
Notations
Use of Exponential Notations in
Addition and Subtraction
Before addition or subtraction,
convert all numbers into the same exponents of 10. Then add or subtract the
digit terms (coefficients)
For example; 1.31 x 103
and 3.15 x 102 can be added using the rule of exponential notation
as
Before addition, the value 3.15 x102
is converted into 0.315 x 103 by placing decimal point to the
left to get same exponents of 10 as that of 1.31 x 103. Then
coefficients of both values are added to get the result.
1.31 x 103
+
0.315 x 103
1.625 x 103
Use of Exponential Notations in
Multiplication
Multiply all digits terms
(coefficients) and add all exponents algebraically. The final result may be
adjusted by placing the decimal to the left or right to get result in standard
scientific notation.
For example; 7.0 x 1012 and 2.0 x 10−3
can be multiplied using the rule of exponential notation as
=
(7.0) (2.0) x 1012−3
= 14 x
109
= 1.4
x 1010
Use of Exponential Notations in
Division
Multiply all digits terms
(coefficients) and subtract all exponents algebraically. The final result may
be adjusted by placing the decimal to the left or right to get result in
standard scientific notation.
For example; 6.60 x 108 divide by 3.20 x 103
can be solved using the rule of exponential notation as
6.60 x 108/3.20 x 103 = 2.60 x 108−3 = 2.60 x 105
Use of Exponential Notations in
Powers
Multiply the digit terms as well as
exponent with a number which indicates the power. The final result may be
adjusted by placing the decimal to the left or right to get result in standard
scientific notation
For example; (3.25 x 104)2 can be solved
using the rule of exponential notation as
Here digit term is 3.25 and
exponent is 104 and both are multiplied by whole power of the figure
i.e. 2 to get the answer
(3.25 )2 x 104x2
= 10.56 x 108 = 1.056 x 109
Use of Exponential Notations in
Roots
Adjust the exponents by shifting the decimal to the right to left in digit term in such a way that exponent must exactly be divided by the root. Then simply the root of digit term and divide the exponent by a desired root.
1.4 Mole
In routine work, we generally measure things by weighing
or by counting with the option based on our comfort. It is more convenient to
buy gloves by pairs (two gloves), bananas by dozens (one dozen is equal to 12
bananas), tea sachets by gross (one gross is equal to 122 sachets) and paper by
reams (one ream is equal to 500 sheets) but purchasing rice from a grocery shop
is convenient by weighing instead of counting.
Similarly, if we prepare a solution
or perform a reaction in chemistry laboratory or even in process in industry we
deal with the enormous number of atoms, molecules, formula units or ion that
mix with one another in specific ratio. Since atom or molecule is so tiny, how
can it be possible to count them? To solve this difficulty, chemistry have
devised a special counting unit called as mole as convenient way to count the
number of particles in chemical substance by weighing them.
Mole (Latin; heap or pile)
is the SI unit for the amount of substance of specific number of particles. A
mole is the gram atomic mass or gram molecular or gram formula mass of any substance
(element or covalent compound or ionic compound) which contains 6.02 x 1023
particles (atoms, molecules, ions etc.).
Mole represent the number of
chemical entities in a fixed mass.
Example
1 mole of hydrogen atom (H) = 1 g =
6.02 x 1023 atoms
1 mole of hydrogen molecule (H2)
= 2 g =
6.02 x 1023 molecules
1 mole of water (H2O)
= 18 g = 6.02 x 1023 molecules
1 mole of sodium chloride (Na+Cl−)
= 58.5 g= 6.02 x 1023 + 6.02
x 1023 ions
1.5 Avogadro’s number
One mole of any substance always
contains 6.02 x 1023 particles and known as Avogadro’s number (NA)
in the honour of Italian physicist Amedeo Avogadro.
The number of particles (atoms,
ions or molecules) present in one mole of any substance (element or compound) is
called Avogadro’s number (NA) and its numerical value if 6.02 x 1023.
One mole of any substance contains
6.02 x 1023 particles and no matter what the chemical nature of a
substance is. Hence one mole of carbon and one mole of magnesium contains same
number of atoms but one mole of magnesium has a mass twice (24g) as that of one
mole of carbon (12g).
Nature of number of particles in
Ionic compounds
Ionic compound is assumed to
consists of formula units. Thus one of mole MgCl2 contains 6.02 x 1023
formula units. But one mole of MgCl2 contains one mole of Mg2+
ion and two moles of Cl− ions. This indicates that one mole of
MgCl2 contains 6.02 x 1023 Mg2+ ions and 12.04
x 1023 (2 x 6.02 x 1023) Cl− ions.
MgCl2
→ Mg2+ + 2Cl−
Relation between mole, molar mass and Avogadro’s number
1.6 Molar Mass
The
mass in grams of one mole of any pure substance is known as molar mass. It is
expressed in grams per mole (g/mol).
Comparing a dozen of bananas and a
mole of carbon, it is important to note that bananas vary in mass but all
carbon atoms have equal mass. Thus a fruit seller cannot sale one dozen bananas
by weighing them, however a chemist can deal with 1 mole of carbon easily by
weighing 12 g carbon.
The basic difference between the
mass of one atom and the mass of 1 mole is that the atomic mass of one atom of
an element is specified by amu and the mass of one mole of an element is
expressed in grams. Thus atomic mass of Fe is 55.85 amu but its molar mass is
55.85 gram. A similar relationship holds for compounds. e.g. molecular mass of
propane (C3H8) is 44 amu but its molar mass is taken as
44 grams.
1.7 Molar Volume
Definition
The volume of one mole of a gas at
standard temperature of 273 K and pressure of 1 atm is known as molar volume
(VM). Molar volume of all ideal gases at STP is 22.4 dm3. Molar
volume is obtained by dividing molar mass of gas with its mass density.
Molar volume
= Molar mass/density
Inter conversion of Mole and Mass
To convert mole into mass and
vice-versa, molar mass is used as conversion factor. Given mass of substance
can be converted into mole by dividing it by its molar mass. Conversely, given
mole of a substance can be converted into its mass by multiplying it by its
molar mass. Conclusively, the following formula may be used for the inter
conversion of mole and mass:
No.
of moles = Given mass (g)/molar mass of substance (gmol−1)
Inter conversion of Mole and Number
of Particles (NP)
To convert mole into number of
particles or vice-versa, Avogadro’s number (NA) is used as
conversion factor. Following formula is used for the inter conversion of mole
and number of particles:
No.
of moles = Given no. of particles of the substance (NP) / Avogadro’s
number (NA)
Inter conversion of Mole and Volume
of gas (Vg) at STP
To convert mole into volume of gas at STP or vice-versa, molar volume (VM) is used as conversion factor. Following formula is used for the inter conversion of mole and volume of gas at STP:
No. of moles = Given volume of gas at STP (Vg) / Molar volume (VM)
1.8 Stoichiometry
Etymology
(Meaning of the Word)
The word
stoichiometry derives from two Greek words; stoicheion meaning “element”
and metron meaning
“measurement”.
Definition
The branch of chemistry which deals with the study of quantitative
relationships between the amounts of reactants and products of a chemical
reaction as given by a balanced chemical equation is called Stoichiometry. It is a very mathematical part of chemistry.
Stoichiometry tells us what amount of each reacting
species we required to consume completely into desired amount of product(s).
Stoichiometric Amounts
The amount of reactants and
products in balanced chemical equations are called Stoichiometric Amounts. According
to Law of Conservation of Mass, the weight of reactants is equal to weight of
products.
Explanation
In stoichiometric calculation of
chemical reactions, two assumptions are made:
(a) Reactants
are completely converted to products.
(b) No
side-reactions occur.
In fact, many chemical reactions are reversible to some
extent, further, in some chemical processes side reaction may also occur.
Stoichiometric Relationships
The stoichiometric calculations
from the balanced chemical equations involve three types of relationships:
1. Mass-Mass
Relationship.
2. Mass-Volume
Relationship.
3. Volume-Volume
Relationship.
1. Mass–Mass
Relationship/Calculation based on Mass–Mass Relationship
Such relationships are helpful in
determining unknown mass of a reactant or product from the given mass of
reactant or product with the help of balanced chemical equation.
2. Mass–Volume Relationship/Calculation
based on Mass–Volume Relationship
The Mass–Volume relationship or
volume-mass calculations are used when either one of the reactant or product is
in gaseous state. The mass – volume
relationships are useful is determining the unknown volume or mass of reactants
or products from a known volume or mass of some substances in a chemical
reaction. It is based on Avogadro’s law which states that 1 mole (1 g/mole) of
every gas always occupies 22.4 dm3 (22.4 litre) or 22400 cm3
or 0.0224 m3 at S.T.P.
3. Volume
– Volume Relationship/Calculation based on Volume – Volume Relationship
Volume-Volume Relationship is used
for determination of volumes of gases. The volume-volume relationships are
useful in determining unknown volume of reactants or products from a known
volume of reactants or products.
It is
based on Gay-Lussac’s Law of Combining Volumes which states that:
“Gases combine or form in
chemical reactions in the ratio of simple whole numbers by volumes provided at
same temperature and pressure”. The ratio of the volumes of gases is same as
the ratios of their molecules in a balanced equation
e.g.
1.9 Limiting Reactant
Definition of Limiting Reactants
The reactant which is entirely
consumed first during chemical reaction is called limiting reactant or limiting
reagent. Mathematically, limiting reactant is that which gives least number of
moles of product
Definition of Excess Reactant
The
reactant that is not completely consumed is called as excess reactant.
Applications of Limiting Reactants
We know that a sandwich is made of
two slices of bread and one shami. Suppose we have 20 slices of bread and 8
shamies. From this quantity we will be able to make only 8 sandwiches. Thus, 4
slices are left over. The available quantity of shamies limits the number of
sandwiches.
The similar situation occurs in
many irreversible chemical reactions where one reactant is often completely
used while some amount of other reactant remains unreacted. In the chemical
reaction because once one of the reactants is consumed completely, the reaction
stops.
1. During
experiments, the stoichiometric amounts of reactants are not usually used.
2. If reactants are not mixed with
stoichiometric amounts then at completion of reaction some reactants remains
unreacted. One of the reactants may be consumed earlier. This reactant is
called Limiting Reactant.
3. Amount of product formed in a chemical
reaction depends upon the amount of limiting reactant.
4. If limiting reactant is completely used,
no further product is formed.
For example
Consider the formation of water
form hydrogen gas (H2) and oxygen gas (O2) in the
following balanced equation:
2H2(g)
+ O2(g) → 2HO(l)
If we start the reaction by taking
2 moles of hydrogen and 2 moles of oxygen, it will be noted that hydrogen is
consumed earlier and stops the reaction while O2 is left behind.
Thus, H2 is considered as a limiting reactant and O2 is
referred as excess reactant.
Steps for identification of Limiting Reactant
To find out a limiting reactant in a chemical reaction we
must focus on following four steps:
1. Write a
balanced chemical equation of the given chemical reaction
2. Determine
the number of moles of reactants from their given amount.
3. The number of moles of required product
is calculated from moles of all reactants.
4. Reactant producing least moles of
product is limiting reactant.
Yield of Reaction or Reaction
Yield
Yield
The amount of the product obtained
as a result of the chemical reaction is called yield.
Types of yield
There are
three types of yields
1. Theoretical Yield
2. Actual
yield
3. Percentage
yield
Theoretical
Yield/ stoichiometric yield/ calculated yield
The maximum amount of the product
calculated from the balanced chemical equation by using its limiting reactant
(given amount of reactant) is known as theoretical yield or stoichiometric
yield or calculated yield or expected yield.
It is the maximum amount of the
product that can be produced theoretically and stoichiometrically by known
amount of a reactant according to balanced chemical equation.
theoretical yield of a
reaction is always greater than the actual yield of the same reaction.
Actual
Yield/ experimental yield
The actual amount of product which
is formed in an experiment is called practical or actual yield. The amount of the products actually
obtained from a given amount of the reactant in a chemical reaction
experimentally is called the actual yield or experimental yield of that
reaction.
The actual yield of a reaction is
always less than the theoretical yield
Percentage
Yield
The efficiency of a chemical
reaction can be checked by calculating its percentage yield which is expressed
by comparing the actual yield and theoretical yields. The ratio of practical
yield to theoretical yield is referred as percent yield
Reason of Less Actual Yield
No matter how much expertise we
have in doing chemical reaction or we use highly efficient techniques, we will
lose some amount of product during the course of reaction in laboratory or
chemical plant. Hence what we actually obtain is less than what we calculate
(theoretical yield).
The actual yield of a reaction is
always less than the theoretical yield because of following reasons:
1. Either
some amount of reactant may not react. The reaction may not go to completion
and may reduce the yield of
the product.
2. The reactant may form any side product.
Side reactions may produce by-products (i.e. some of the reactants might take
part in a competing side reaction and reduced the amount of the desired
product. Competing side reactions may also occur decreasing practical yield.
3. Reversibility
of reactions leads to less products.
4. Materials
may be lost in handling.
5. Mechanical
loss of certain amount of product takes place due to filtration, distillation
and separation by separating
funnel, washing and crystallization etc.
6. A
practically inexperienced worker has many shortcomings and cannot get the
expected yield.
Numericals on Theoretical and Percentage Yield
Q2. The reaction of calcium carbonate (CaCO3)
with hydrochloric acid is given as
CaCO3(s)
+ 2HCl(aq) → CaCl2 + CO2(g) + H2O(l)
If during an experiment 50 g of CaCO3 is reacted with
excess of hydrochloric acid, 14.52 g of CO2 is liberated, calculate the theoretical and percentage
yield of CO2 gas.
Solution
(I) Conversion
of mass of CaCO3 into moles
Moles
of CaCO3 = mass/molar mass = 50/40+12+48 = 50/100 = 0.5 moles
(II) Calculation
of molar amount of product from molar amount of reactant
CaCO3(s)
+ 2HCl(aq) → CaCl2 + CO2(g) + H2O(l)
(Given) (required)
1 mole
----------------------------> 1
mole
0.5 mole
--------------------------> ?
mole
According
to unitary method from balanced chemical equation
1 mole
of CaCO3 gives 1 mole of CO2
0.5
mole of CaCO3 gives = 0.5 mole of CO2
(III) Calculation
of Theoretical Yield
This is
the theoretical yield in mole. Now it is converted into to theoretical yield by
multiplying with molar mass
Theoretical
yield = moles x molar mass = 0.5 x 12+32 = 0.5 x 44 = 22 g
(IV) Calculation
of Percentage Yield
Percentage
yield = Practical or actual yield /theoretical yield x 100
= 14.52/22 x 100 = 66%
Q3. Under high pressure magnesium (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO).
2Mg + O2 → 2MgO
If 4 grams of Mg reacts with excess of O2 to produce 4.24 g of MgO, calculate the percentage yield of MgO.
Solution
(I) Conversion
of mass of Mg into moles
Moles
of Mg = mass/molar mass = 4/24 = 0.166 moles
(II) Calculation
of molar amount of product from molar amount of reactant
2Mg + O2 →
2MgO
(Given) (required)
2 mole
---------> 2 mole
0.166 mole -----> 0.5
mole
According
to unitary method from balanced chemical equation
2 mole
of Mg gives 2 mole of MgO
= 0.166
mole of MgO
(III) Calculation
of Theoretical Yield
This is
the theoretical yield in mole. Now it is converted into to theoretical yield by
multiplying with molar mass
Theoretical yield = moles x molar mass
Theoretical yield = 0.166 x 24+16
Theoretical yield = 0.166 x 40
Theoretical yield = 6.64 g
(IV) Calculation
of Percentage Yield
Percentage
yield = Practical or actual yield /theoretical yield x 100
Percentage yield = 4.24/6.64 x 100
Percentage yield = 63.85%
Q3. Aluminium chloride is used in the
manufacturing of rubber. It is produced by allowing aluminium to react with Cl2 gas at 650oC.
2Al(s) + 3Cl2(g) → 2AlCl3(l)
When 160 g aluminium reacts with excess of chlorine, 650 g of AlCl3 is produced. What is the percentage yield of AlCl3?
[Ans: 82.17%]
Q4. Potassium chlorate decomposes upon slight
heating in the presence of a catalyst, according
to the reaction below.
2KClO3(s) → 2KCl(s) + 3O2(g)
In a certain experiment, 40.0g KClO3 is heated until it completely decomposes. What is the theoretical yield of oxygen gas? The experiment is performed, the oxygen gas is collected, and its mass is found to be 14.9 g. What is the percent yield for the reaction?
Solution
First, we will calculate the theoretical yield based on the
stoichiometry and then the percent yield.
(I) Conversion
of mass of KClO3 into moles
Mass of KClO3 = 40.0g
Molar mass KClO3 = 39 + 35.5 + 48 = 122.55
g/mol
Molar mass O2 = 16 x 2 = 32.00g/mol
Mole ratio = 3/2 or 3 moles of O2/2 moles of KClO3
Moles
of KClO3 = mass/molar mass = 40.0/122.5 = 0.0326 mol
(II) Calculation
of molar amount of product from molar amount of reactant
2KClO3(s) → 2KCl(s) + 3O2(g)
(Given) (required)
2 mole
-----------------> 3 mole
0.0326 mole
---------> ? mole
According
to unitary method from balanced chemical equation
2 mole
of 2KClO3 gives 3 mole of O2
0.0326 mole of 2KClO3 gives = 3/2 x 0.0326 mole of O2
= 0.489 mole
of O2
Alternate
Method
Desired
quantity = Given quantity x conversion factor
= 0.0326
x 3/2 = 0.489 mol O2
(III) Calculation
of Theoretical Yield
This is
the theoretical yield in mole. Now it is converted into to theoretical yield by
multiplying with molar mass
Theoretical
yield = moles x molar mass = 0.489 x 32 = 15.684 g O2
The theoretical yield of O2 is 15.7g
(IV) Calculation
of Percentage Yield
Actual yield =14.9g
Theoretical yield =15.7g
Percentage
yield = Practical or actual yield /theoretical yield x 100
= 14.9/15.684 × 100% = 94.9%
Q5. For the
reaction; Be + 2 HCl → BeCl2 + H2, the theoretical yield
of beryllium chloride was 10.7 grams. If the reaction actually yields 4.5
grams, what was the percent yield?
Solution
% 𝑌𝑖𝑒𝑙𝑑 = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑/𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 × 100% = 4.5 𝑔 /10.7 𝑔 × 100% = 42%
Q6. If the percentage yield is 45% with the
theoretical yield as 4g, what would the actual yield be? Calculate using the
percentage yield formula.
Solution
Using the percentage yield formula,
Percentage yield
= (Actual yield/Theoretical yield)× 100%
45 = Actual
yield/4 × 100
Actual yield =
1.8g
Therefore, the
actual yield is 1.8g
Assignment
Numericals
Q1. Phosphorous reacts with bromine to form phosphorous
tribromide. If 35.0 grams of bromine are reacted and 27.9 grams of phosphorous tribromide
are formed, what is the percent yield? 2P + 3Br2 → 2PBr3 (Answer;
70.63%)
Q2. Silver Nitrate reacts with Magnesium
Chloride to produce Silver Chloride and Magnesium Nitrate. If 305 grams of
silver nitrate are reacted in an excess of magnesium chloride producing 23.7
grams of magnesium nitrate, what is the percent yield? (Answer; 8.91%)
Q3. If the reaction of 6.5 grams of C6H12O6
produces 2.5 grams of CO2, what is the percent yield?
C6H12O6 →
2C2H5OH
+ 2CO2
Q4. If the reaction of 125 grams of C6H6O3
reacts in excess of oxygen (O2) and produces 51 grams of H2O,
what is the percent yield?
C6H12O3 + 3O2 →
6CO + 3H2O
Q5. During a chemical reaction, 0.5 g of
product is made. The maximum calculated yield is 1.6 g. What is the percent
yield of this reaction? (Answer; )
Q6. During a chemical reaction 1.8 g of
product is made. The maximum calculated yield is 3.6 g. What is the percent
yield of this reaction?
Q7. If the percentage yield is 45% with the
theoretical yield as 4g, what would the actual yield be? Calculate using the
percentage yield formula.
Q8. During a chemical reaction, 0.9 g of
product is made. The maximum calculated yield is 1.8g. Calculate the percentage
yield of this reaction by using the percent yield formula chemistry?
Q9. Determine the theoretical yield of the
formation of geranyl formate from 465 g of geraniol. A chemist making geranyl
formate uses 465 g of starting material and collects 419g of purified product.
Percentage yield is given as 92.1%.
Q10. What is the percent yield of the following
reaction if 60 grams of CaCO3 is heated to give 15 grams of
CaO?
Q11. Calculate the percent yield of sodium
sulfate when 32.18 g of sulfuric acid reacts with excess sodium hydroxide to
produce 37.91 g of sodium sulfate.
No comments:
Post a Comment