Quantitative Chemistry

Question 1

What is the law of
conservation of mass?

Answer 1

The law of conservation of mass states that no atoms are lost or made
during a chemical reaction so the mass of the products equals the mass
of the reactants.

Question 2

Write a balanced equation
of magnesium reacting with
hydrochloric acid.

Answer 2

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

Question 3

Define relative atomic mass
and relative formula mass.

Answer 3

RAM - average mass of atoms in an element taking into account masses
and abundance of its isotopes, relative to 12C.
RFM - sum of RAM’s of all atoms in the formula.

Question 4

What is the relative formula
mass of:
A) CaF2
B) C6H12O6

Answer 4

CaF2
- (Ar
values: Ca = 40, F = 19)
40 + 19 + 19 = 78
—
C2H12O6
- (Ar
values: C = 12, H = 1, O = 16)
(12 x 6) + (1 x 12) + (16 x 6) = 180

Question 5

The following reaction occurs in a test tube under a Bunsen
Burner:
4 MgO(s) + CH4
(g) → 4 Mg(s) + 2 H2O(g) + CO2
(g)
The carbon dioxide and water escape from the test tube.
Use the equation to explain why.

Answer 5

They are both gases

Question 6

Mass of magnesium oxide used
in g
4.0 4.0 4.0
Mass of magnesium produced
in g
3.3 3.5 3.2
The experiment was repeated three
times. Calculate the mean mass of
magnesium produced and suggest
how you could increase the precision
of the results.

Answer 6

(3.3 + 3.5 + 3.2) / 3 = 3.3
Measure to more decimal places or use a more sensitive balance /
apparatus

Question 7

What is Avogadro’s
constant?

Answer 7

The number of atoms, molecules or ions in a mole of a given substance.
The value of the constant is 6.02 x 1023
.

Question 8

What is the formula that
links mass, molecular mass
and moles together

Answer 8

Mass = Mr x Moles

Question 9

What is the mass of:
20 moles of calcium
carbonate, CaCO3

Answer 9

Mass = Mr x Moles
Mr = 100
100 x 20 = 2000 g

Question 10

Calculate the amount of carbon dioxide
in moles in 0.32 g of carbon dioxide.
Relative atomic masses (A
r
): carbon =
12, oxygen = 16

Answer 10

Moles = Mass / Mr
0.32 / 44 = 0.007

Question 11

Nitrogen and hydrogen form ammonia shown by the following
equation:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
Calculate the mass of nitrogen needed to form 6.8 tonnes of
ammonia.
Relative atomic masses (A
r
): H = 1; N = 14

Answer 11

Step 1 - Work out the number of number of moles of ammonia (Mr of ammonia = 17)
6800000 / 17 = 400000 moles of ammonia
Step 2 - Use the balanced equation and number of moles of ammonia to work out the number of moles of
nitrogen
The ratio of nitrogen to ammonia is 1:2
Therefore the number of moles of nitrogen is 400000/2 = 200000
Step 3 - Work out the mass of nitrogen (Mr of N2
is 28)
200000 x 28 = 5600000 g = 5.6 tonnes.

Question 12

State what we mean by a
limiting reactant in a
chemical reaction

Answer 12

In a chemical reaction involving two reactants, it is common to use an
excess of one of the reactants to ensure that all of the other reactant is
used. The reactant that is completely used up is called the limiting
reactant because it limits the amount of products

Question 13

Hydrogen peroxide decomposes in
water to form water and oxygen. How
many grams of oxygen gas will be
given off from 40.8 g of hydrogen
peroxide?

Answer 13

Step 1: Write the balanced equation 2 H2O2(l) → 2 H2O + O2(g) Mr of H2O2 = 34 Step 2: Number of moles in 40.8 g : 40.8/34 = 1.2 molesRatio in the balanced equation of H2O2
: O2
= 2:1
Step 3 :Therefore number of moles of O2
= 0.6 moles
Step 4: Mass of oxygen = 0.6 x 32 (Mr of O2
) = 19.2

Question 14

Write down the two formulae that
link concentration, mass and
volume together.

Answer 14

Concentration (g per dm3
) = Mass (g)/Volume (dm3
)
Concentration (mol per dm3
) = nr of moles/volume (dm3
)

Question 15

31.0 cm3
of potassium hydroxide solution
neutralised 25.0 cm3
of 2.0 moldm−3 nitric acid.
HNO3
+ KOH → KNO3 + H2O
Calculate the concentration of the potassium
hydroxide solution in moldm−3

Answer 15

Step 1: Calculate the moles of HNO3
used = Concentration x volume
2 x 0.025 dm3
(25/1000 to convert the units) = 0.05 moles
Step 2 : Calculate the moles of KOH
Ratio is 1:1 therefore number of moles of KOH = 0.05
Step 3 : Calculate the concentration of KOH
Volume = Moles/concentration; 0.05 / 0.031 = 1.61

Question 16

What is the molar volume of a gas
at room temperature and
pressure?

Answer 16

1 mole of a gas at room temperature and pressure occupies 24 dm3

Question 17

What is titration?

Answer 17

A technique for finding the concentration of a solution by reacting a known
volume of this solution with a solution of known concentration.

Question 18

How do you conduct a titration?

Answer 18

a) Rinse the pipette with a solution of unknown concentration. Use the pipette to
measure out the known volume of this solution.
b) Add an indicator (a substance that changes colour at the end of titration)
c) Rinse the burette with a solution of known concentration. Discard the liquid.
Use a burette to gradually add the solution of a known concentration.
d) When indicator changes colour (at the end point), the volume added is
recorded
e) It is important to get concordant volume results - they have to lie close to each
other
f) Suitable calculations are performed to find the concentration.

Question 19

Why is it not always possible to
obtain the theoretical amount of
product in a chemical reaction?

Answer 19

● The reaction may not go to completion because it is reversible.
● Some of the product may be lost when it is separated from the
reaction mixture.
● Some of the reactants may react in ways different to the expected
reaction (side reactions may occur).

Question 20

How is the percentage yield of
a product in a chemical
reaction?

Answer 20

Yield = Actual mass of a product x 100% /
Maximum theoretical mass of product

Question 21

What is the % yield of NH3
if 40.5 g
NH3
is produced from 20.0 mol H2
and excess N2
?

Answer 21

Step 1 - Write the balanced equation
N2
+ 3 H2
→ 2 NH3
Step 2 - Calculate the theoretical amount of NH3 .Moles NH3
(ratio of H2
to NH3 is 3:2);
of 20/1.5 = 13.3 moles
13.3 X 17 (Mr of NH3
) = 227
Step 3 - Calculate percentage yield of NH3
40.5/227 x 100 = 17.8%

Question 22

What is atom economy?

Answer 22

A measure of the amount of starting materials that end up as useful products.
It is a ratio of the relative formula mass of desired product to the sum of relative
formula masses of reactants.