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# Percentage Composition - Definition, Examples, Practice Problems & Frequently Asked Questions-FAQs

How many of you live in Delhi? Or belong to Maharashtra?
How many of you practise Sikhism? Or follow Jainism?
Our country is vastly diverse on many parameters ranging from languages, religions, clothes, races, food, tribes etc.
It would be a very difficult scenario to know for certain the framework of every individual and to correlate the same with respect to the entire country’s population.
If I was to ask you how many people are below the age of 21 in relation to people who are under 50?
Or the arrangement of the total number of Bengali people with respect to people who belong to Rajasthan?
Such questions and their answers drive us towards the concept of “composition”.
So, let’s study the concept of “percentage composition” and try to understand it.

## Percentage composition

In 100 parts of the compound, it is the relative mass of each of the constituent elements.

The mass percent of an element =

Calculation of the percentage of elements present in a molecule is a straightforward task if the molecular formula is known.

According to the law of definite proportions, a compound mass ratio of elements is always the same irrespective of their source or mode of manufacturing.

1 mole of H2O always contain 2 mol of hydrogen and 1 mole of oxygen means 18 g of H2O made up of 2 g hydrogen and 16 g oxygen.

Molar mass of H2O=18 g mol-1

## Molar mass of gaseous mixture

We know our atmospheric air is a mixture of nitrogen gas (78 %), oxygen gas (21 %), argon gas (0.93 %), carbon dioxide gas (0.04 %) and traces of many gases like water vapour, hydrogen, helium, etc.

For numerical simplicity, treat air is a mixture of three gases N2, O2 having 78 %, 22 % by volume.

So, % of N2 gas = 78%

% of O2 gas = 22%

We can conserve mass;

Mass of the mixture (air) = mass of N2 gas + mass of O2 gas

We know, d=mV and PM = dRT, d=PMRT

(d = density, m = mass, V = volume, M = molar mass, R = universal gas constant)

and T = temperature)

dmixtureVmixture=dN2VN2+dO2VO2

dmixtureVmixture=dN2VN2+dO2VO2

dmixtureVmixture=dN2VN2+dO2VO2

PMmixtureRTVmixture=PMN2RTVN2+PMO2RTVO2

At a constant Temperature (T) and Pressure (P)

MmixtureVmixture=MN2VN2+MO2VO2

Mmixture=MN2VN2+MO2VO2Vmixture

Mmixture=(28)(78)+(32)(22)100=28.88 g mol-1

## Calculation of minimum molecular mass

For a unknown compound, if at least % of one element is given then we can find the minimum molecular mass of that element. At least one atom of the given element is present in one molecule of the sample.

E.g - During an experiment, a compound contains 0.6 % iron by mass. Calculate the minimum molecular mass of the compound?

Answer: Let the molecular mass of the compound = x amu

At least 1 atom of Fe is present in that compound

We know the atomic mass of Fe = 56 amu

Given, mass % of Fe = 0.6 %

According to question, 56 amu is the 0.6 % of x amu.

x0.6100=56

x=9333.33 amu

## Practice problems

Q1. Find the mass percent of oxygen atom in sulphuric acid

• 65.31 %
• 27.09 %
• 32.65 %
• 48.98 %

Solution: Molecular formula of sulphuric acid = H2SO4

Molar mass of H2SO4= 98 g mol-1

Mass of oxygen = 416 g= 64 g

% mass of O - atom = mass of oxygenmass of H2SO4100=6498100=65.31 %

Q2. A sample of chlorophyll was studied and the research team analyzed and concluded that chlorophyll contains 0.066% magnesium. If each molecule of chlorophyll is supposed to have 2 atoms of magnesium, calculate the molecular weight of the chlorophyll?

• 18181.82 amu
• 72727.27 amu
• 36363.64 amu
• None of these

Solution: Let the molecular mass of chlorophyll = x amu

At least 1 atom of Mg is present in that chlorophyll

We know the atomic mass of Mg = 24 amu

Given, mass % of Mg = 0.066 %

According to question 2 atoms of Mg is present per molecule of chlorophyll, 48 amu is the 0.066 % of x amu.

x0.066100=48 amu

x=72727.27 amu

Q3. A chemical compound is a molecule with 21 oxygen atoms. the compound has 71 % oxygen by mass. Its molar mass is

• 177.46 g
• 532.39 g
• 106.48 g
• 473.24 g

Solution: Let the molar mass of compound = x g

Given, mass % of oxygen = 71 %

According to question 21 atoms of oxygen atoms are present per molecule of the compound,

Molar mass of 21 O atoms = 336 g

336 g is the 71 % of x g.

x71100=336 g

x=473.24 g

Q4. Which has the highest mass percentage of nitrogen

• NH3
• NO2
• N2O
• N2O5

Solution:

Molar mass of NH3= 17 g mol-1

Molar mass of NO2= 46 g mol-1

Molar mass of N2O= 44 g mol-1

Molar mass of N2O5= 108 g mol-1

Molar mass of nitrogen = 14 g mol-1

% mass of N - atom in NH3 = molar mass of nitrogenmolar mass of NH3100=1417100=82.35 %

% mass of N - atom in NO2 = molar mass of nitrogenmolar mass of NH3100=1446100=30.43 %

% mass of N - atom in N2O = molar mass of nitrogenmolar mass of NH3100=2844100=63.64 %

% mass of N - atom in N2O5 = molar mass of nitrogenmolar mass of N2O5100=28108100=25.93 %

Question 1. What more applications of percentage composition can we find in our daily lives?
Answer: We can observe many applications of percentages in daily life. E.g- You can find GST % in hotel bills, telephone bills, etc. if you open any website or app which displays the share market you can find information in percentages. Hydrogen peroxide bottles in a medical shop containing 2 % w/V solution, Whisky, beer, wine, etc their alcohol percentage is mentioned on their bottle.

Question 2. What factors are involved in percentage composition?
Answer: for mass percentage, the percent composition of a compound may be calculated by dividing the mass of each element by the total mass of the complex.

We can calculate the percentage of any variable by dividing the factor for single species by the value of the total species.

Question 3. Explain whether or not all pure samples of a specific substance have the same % composition.
Answer: according to definite proportion, a compound mass ratio of their constituents is always the same irrespective of their source or method of manufacturing. So, their % composition will be always the same as a pure compound.

Question 4. How do you estimate the empirical formula based on the % composition?
Answer: It is very easy to calculate the percentage of elements present in a molecule if the molecular formula is known but if the percentage of elements is known then finding the molecular formula is not a simple task.

First, we have to calculate the relative number of atoms of each element in the particular compound. Then, we can find the empirical formula of the compound. From some other technique, we have to find the molecular mass of the compound then we are able to calculate the exact molecular mass of the compound.

Rules to finding empirical formula of any compound if percentage composition is known

Step 1: Relative mole of atoms

If a compound on analysis gave the following results: C = 39.99%, H = 6.71% and O = 53.28%.

 Element Percentage Molar weight (g) No. of moles C 39.99% 12 3.33 H 6.71% 1 6.71 O 53.28% 16 3.33

Step 2: Find the simplest ratio of moles of atoms

 Element Percentage Molar weight (g) No. of moles Simplest whole no. ratio C 39.99% 12 39.9912= 3.33 3.333.33= 1 H 6.71% 1 6.711= 6.71 6.713.33= 2 O 53.28% 16 53.2816= 3.33 3.333.33= 1

Thus, CH2O this is the empirical formula of the above compound.

Related topics

 Empirical Formula Atomic Number and Mass Number mole Atomic mass Average atomic mass Law of chemical reaction
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