Quantitative Chemistry

12 methods · 29 questions

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Turn every element into moles by dividing its mass (or its percentage read as a mass out of 100) by its $A_{r}$ , then divide through by the smallest amount to get the atom ratio; that ratio is the empirical formula. For a molecular formula, scale the whole empirical unit by $\frac{M _{r} ( molecule )}{M _{r} ( empirical unit )}$ .

Trigger: a compound described by the masses or mass fractions of its elements, or a molecular mass stated as a multiple of the empirical mass.

Instances:
(i) mass data: subtract to recover any missing element, divide each mass by its $A_{r}$ , then take the simplest ratio (a $2 : 5$ ratio gives $B r_{2} O_{5}$ );
(ii) an awkward ratio such as $1.5 : 2$ is not an error: multiply the whole ratio up to whole numbers ( $1.5 : 2 = 3 : 4$ , giving $F e_{3} O_{4}$ ), never round;
(iii) molecular from empirical: find the empirical mass, then multiply every subscript by $M_{r}$ over that mass ( $C_{2} F_{3}$ has mass $81$ , the molecule $162$ , so $C_{4} F_{6}$ );
(iv) if composition is given as fractions and one element's absolute mass is fixed by the chemistry (bromine adds as $B r_{2}$ , two atoms per molecule), read $M_{r}$ as that element's mass divided by its mass fraction.

Linked questions (3)

NSAA 2016 Section 1, Part C (Chemistry questions) · Q41

Several oxides of bromine have been identified. Analysis of

2.4 g

of one of these compounds showed it to contain

1.6 g

of bromine.

What is the empirical formula of this compound?

(

A_{r}

: bromine = 80; oxygen = 16)

A $B r_{2} O$
B $Br O_{2}$
C $B r_{2} O_{5}$
D $B r_{5} O_{2}$
E $B r_{4} O_{5}$
F $B r_{5} O_{4}$

Worked solution

Answer:C

The compound contains

2.4 - 1.6 = 0.8 g

of oxygen.

Moles of bromine:

\frac{1.6}{80} = 0.02 mol

. Moles of oxygen:

\frac{0.8}{16} = 0.05 mol

.

The ratio Br : O is

0.02 : 0.05 = 2 : 5

, so the empirical formula is

B r_{2} O_{5}

.

The answer is C.

NSAA 2017 Section 1, Part C (Chemistry questions) · Q46

A fluorocarbon has a relative molecular mass which is twice that of its empirical formula mass.

81 g

of the compound contains

57 g

of fluorine.

What is the molecular formula of the compound?

(

A_{r}

values: C = 12; F = 19)

A $C_{2} F_{3}$
B $C_{2} F_{5}$
C $C_{3} F_{6}$
D $C_{3} F_{8}$
E $C_{4} F_{6}$
F $C_{4} F_{10}$

Worked solution

Answer:E

A fluorocarbon contains only carbon and fluorine, so the mass of carbon in the sample is

81 - 57 = 24 g

.

Convert to moles: carbon

24/12 = 2 mol

, fluorine

57/19 = 3 mol

. The simplest whole-number ratio is

C : F = 2 : 3

, so the empirical formula is

C_{2} F_{3}

, with an empirical formula mass of

2 \times 12 + 3 \times 19 = 81

.

The relative molecular mass is twice this,

162

, so the molecular formula contains twice the empirical unit:

C_{4} F_{6}

. As a check,

4 \times 12 + 6 \times 19 = 48 + 114 = 162

.

The answer is E.

NSAA 2019 Section 1, Part C (Chemistry questions) · Q43

116 g

sample of an oxide of iron contains

84 g

of iron.

Which of the following is the empirical formula of this oxide of iron?

(

A_{r}

values:

O = 16

;

Fe = 56

)

A $FeO$
B $F e_{2} O_{2}$
C $F e_{3} O_{2}$
D $F e_{2} O_{3}$
E $F e_{3} O_{4}$

Worked solution

Answer:E

The oxide contains

84 g

of iron, so the mass of oxygen is

116 - 84 = 32 g

.

Converting to moles: iron

= \frac{84}{56} = 1.5 mol

and oxygen

= \frac{32}{16} = 2 mol

.

The ratio of iron atoms to oxygen atoms is therefore

1.5 : 2

. Multiplying by

2

to obtain whole numbers gives

3 : 4

, so the empirical formula is

F e_{3} O_{4}

.

As a check,

F e_{2} O_{3}

would need a ratio of

1 : 1.5

, that is

2.25 mol

of oxygen (

36 g

) for

1.5 mol

of iron, which does not match the

32 g

of oxygen in the sample.

The answer is E.