Percentage Composition

Percentage composition helps in checking the purity of a given sample.

By Sakshi Goel | 28 Oct'18 | 1 K Views |

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Theory

Percentage composition of the compound is the relative mass of each of the constituent element in 100 parts of it.

$M a s s p e r c e n t o f a n e l e m e n t = \frac{m a s s o f t h a t e l e m e n t i n o n e m o l e o f c o m p o u n d}{m o l a r m a s s o f t h e c o m p o u n d} Ã— 100$

For example: For water (H₂O):
Molar mass of water = 18.02 g/mol, Molar mass of H = 1.008, Molar mass of O = 16.00
$M a s s % o f H = \frac{2 Ã— 1.008}{18.02} Ã— 100 = 11.18 %$
$M a s s % o f O = \frac{16.00}{18.02} Ã— 100 = 88.79 %$

Empirical formula and molecular formula:

An empirical formula simply gives the relative number of atoms of each element in a compound.
A molecular formula gives the actual number of atoms of each element in a molecule of a compound.
For example for glucose, C₆H₁₂O₆ is the molecular formula. When we see the ratio of the elements, it is 1:2:1. So, empirical formula is CH₂O. Similarly for B₂H₆, molecular formula is B₂H₆, but empirical formula is BH₃.

Obtaining empirical formula from experimental data:

If we know the percentage composition of each of the constituent elements, we can find the empirical formula of the compound. For example, for a compound, the percentage composition is C = 80%, H=20%. Find the empirical formula.

Element	%	Atomic mass	Mass in sample	Number of moles	Sample ratio	Simplest whole number ratio
C	80	12	80 g	$\frac{80}{12} = 6.66$	$\frac{6.66}{6.66} = 1$	1
H	20	1	20 g	$\frac{20}{1} = 20$	$\frac{20}{6.66} = 3$	3

So, Empirical formula is CH₃.

MCQ

Problem:
What is the mass percent of oxygen in Al₂(SO₄)₃.18H₂O. (M = 666.43 g/mol)
(a) 9.6
(b) 28.8
(c) 43.2
(d) 72

Solution:

Molar mass Al₂(SO₄)₃.18H₂O = 666.43 g/mol
Total number of oxygen atoms in the formula = 30
Total mass of oxygen atoms = 16 x 30 = 480 g/mol
mass percent of oxygen = $\frac{480}{666.43} Ã— 100 = 72 %$
Correct option is (d).

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Problem:
A mixture of pure AgCl and AgBr contains 60.94% Ag by mass. What is % of AgCl in sample?
(a) 80.36 %
(b) 19.64 %
(c) 4.8 %
(d) 34.19 %

Solution:

Consider weight of sample mixture = 1 g
Let mass of AgCl = x g
Mass of AgBr = (1 - x) g
Molar mass of AgCl = 143.5 g/mol
Molar mass of AgBr = 188 g/mol
molar mass of Ag = 108 g/mol
Mass % of Ag in AgCl = $\frac{108}{143.5} Ã— 100 = 75.3 %$
Similarly, mass % Ag in AgBr = $\frac{108}{188} Ã— 100 = 57.4 %$
Total % Ag in mixture = 60.94 %
So, (75.3 % of x) + (57.4 % of (1-x)) = 60.94 % of 1 g
0.753x + 0.574(1-x) = 0.6094
0.753x + 0.574 - 0.574x = 0.6094
0.179x = 0.0354
x = 0.198 g
% AgCl in sample = $\frac{0.198}{1} Ã— 100 = 19.8 %$
Closest option is (b).

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Problem:
10 g of H₂O₂ impure sample when reacted with KI (solution) produced 0.5 g iodine. What is the percentage purity of H₂O₂?
(a) 66.9 %
(b) 0.669 %
(c) 100 %
(d) can't be predicted

Solution:

The reaction is:
$H_{2} O_{2} + 2 KI â†’ I_{2} + 2 KOH$
1 mole of H₂O₂ produce 1 mole of I₂.
Molar mass of H₂O₂= 34 g/mol
mass of H₂O₂= 10 g
Moles of H₂O₂= 10/34 = 0.29 mol
Molar mass of I₂ = 254 g/mol
mass of I₂ = 0.5 g
Moles of I₂ = 0.5/254 = 0.002 mol
0.29 mol of H₂O₂ should produce 0.29 mol of I₂
Purity in H₂O₂ = x %
So, x % of 0.29 = 0.002
$\frac{x}{100} Ã— 0.29 = 0.002$
x = 0.689 %
Closest answer is (b)

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Problem:
Percentage of Se in peroxidase enzyme is 0.5% by weight (Se = 78.4). Then the minimum molecular weight of peroxidase enzyme is
(a) 3.13 x 10⁴
(b) 15.68
(c) 1.568 x 10³
(d) 1.568 x 10⁴

Solution:

Percentage of Se = 0.5% by weight
Atomic mass of Se = 78.4
If molecular weight of enzyme = x g/mol
Even if only one atom of Se is present in enzyme, then,
0.5% of x = 78.4
$\frac{0.5}{100} Ã— x = 78.4$
$x = 78.4 Ã— \frac{100}{0.5}$ = 15680 = 1.568 x 10⁴
Correct option is (d).

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Problem:
1.24 g â€˜Pâ€™ is present in 2.2 g of
(a) P₂S₄
(b) P₂S₂
(c) P₄S₃
(d) PS₂

Solution:

If 1.24 g of P is present in 2.2 g of compound, % of P = (1.24/2.2) x 100 = 56.4 %
In option (a), P₂S₄, molar mass = 190 g/mol, total P present = 62 g/mol
% P = (62/190) x 100 = 32.6 %
In option (b), P₂S₂, molar mass = 126 g/mol, total P present = 62 g/mol
% P = (62/126) x 100 = 49.2 %
In option (c), P₄S₃, molar mass = 220 g/mol, total P present = 124 g/mol
% P = (124/220) x 100 = 56.4 %
In option (d), PS₂, molar mass = 95 g/mol, total P present = 31 g/mol
% P = (31/95) x 100 = 32.6 %
Correct option is (c)

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Problem:
The simplest formula of a compound containing 50 % of element X (atomic weight = 10) and 50 % of element Y (atomic weight = 20) is
(a) XY
(b) X₂Y
(c) XY₂
(d) X₂Y₃

Solution:

To have both as 50 %, the total mass of both X and Y should be equal. We will consider the options here:
In XY, total mass of X = 10 g, total mass of Y = 20 g. So, this option is incorrect.
In X₂Y, total mass of X = 2 x 10 = 20 g, total mass of Y = 20 g. So, both X and Y are 50 % each. This option is correct.
In XY₂, total mass of X = 10 g, total mass of Y = 2 x 20 = 40 g. So, this option is incorrect.
In X₂Y₃, total mass of X = 2 x 10 = 20 g, total mass of Y = 3 x 20 = 60 g. So, this option is incorrect.
Option (b) is correct.

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Short Answer

Problem:
An element, X has the following isotopic composition ²⁰⁰X : 90%, ¹⁹⁹X : 8%, and ²⁰²X : 2.0%. What is the weighted average atomic mass of the naturally occurring element X?

Solution:

$A v e r a g e a t o m i c m a s s = \frac{{âˆ‘}_{}^{} p e r c e n t a g e Ã— a t o m i c m a s s}{100}$

$= \frac{(200 Ã— 90) + (199 Ã— 8) + (202 Ã— 2.0)}{100}$

= 200 amu

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