Questions on empirical formula and molecular formula with Answers

Solving Empirical Formula Problems There are two common types of empirical formula problems. Luckily, the steps to solve either are almost exactly the same. Example #1: Given mass % of elements in a compound.

A compound was found to contain 32.65% Sulfur, 65.3% Oxygen and 2.04% Hydrogen. What is the empirical formula of the compound?

1) The first step in this problem is to change the % to grams.
32.65%→32.65g of S
65.3%→65.3g of O
2.04%→2.04g of H

2) Next divide all the given masses by their molar mass.
32.65g of S/ 32gm-1 = 1.0203 moles of S
65.3g of O/ 16gm-1 = 4.08 moles of O
2.04g of H/ 1.008gm-1 = 2.024 moles of H

3) Then, pick the smallest answer in moles from the previous step and divide all the answers by that. Remember that if you calculate a number that is x0.9 round to the nearest whole number
1.0203 moles of S/ 1.0203 = 1
4.08 moles of O/1.0203 = 3.998 ≈ 4
2.024 moles of H/1.0203 = 1.984 ≈ 2

4) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.
S = 1
O = 4
H = 2

H2SO4

Example#2: Given the mass of a reactant before a chemical reaction and the mass of a product after a reaction.

When 0.273g of Mg is heated in a Nitrogen (N2) environment a chemical reaction occurs. The product of the reaction is 0.378g . Calculate the empirical formula.

1) In any empirical formula problem you must first find the mass % of the elements in the compound. Since the total mass of the final product was 0.378 we find that:

0.378g total-0.273g magnesium = 0.105g nitrogen

0.105g nitrogen/0.378g total (100) = 27.77%

0.273g magnesium/0.378g total (100) =72.23%

2) Then change the % to grams
27.77%→27.77g of N
72.23%→72.23g of Mg

3) Next, divide all the masses by their respective molar masses.
27.77g/14gm-1 = 1.98 moles N
72.23g/24.31gm-1 = 2.97 moles Mg

4) Pick the smallest answer of moles and divide all figures by that.
1.98 moles N/ 1.98 = 1
2.97 moles Mg/ 1.98 = 1.5

Since our answer for magnesium is not close enough to round to the nearest whole number we must choose a factor to multiply all the figures by that will yield us whole numbers
1*2= 2 N
1.5*2= 3 Mg

5) Lastly, the coefficients calculated in the previous step will become the subscripts in the chemical formula.
1.98 moles N/ 1.98 = 1
1.98 moles N/ 1.98 = 1
N = 2
1.98 moles N/ 1.98 = 1
Mg = 3

Mg3N2

EMPIRICAL AND MOLECULAR FORMULAE

EMRIRICAL FORMULA – This represents the simplest whole number ratio of

atoms within a molecule

MOLECULAR FORMULA – This represents the exact numbers of

each atom in a molecule

Empirical formulae can be calculated from either % mass compostion data or

mass composition data. Molecular formula can then be calculated if the molecular

mass of the compound is known.

EXAMPLE 1 – from % composition

A substance containing carbon and hydrogen and oxygen only contains 73.47%

carbon and 10.20% hydrogen by mass, its relative molecular mass is 98. Calculate its

molecular formula.

Step 1 calculate empirical formula

ATOMSCARBONHYDROGENOXYGEN

PERCENTAGE MASS73.4710.20(100 – (73.47 + 10.20) =

16.33

ATOMIC MASS12116

MASS/ATOMIC MASS6.122510.201.020

RATIO(divide by

smallest)

6101

Empirical formula = C6H10O

Step 2 use MR to find molecular formula

MR = mass (C6H10O) x n

98 = (98) x nn = 1molecular formula =

C6H10O

EXAMPLE 2 – from mass data

A substance contains 12.28g nitrogen, 3.51g hydrogen, 28.07g sulphur and

56.10g oxygen, its MR is 228 .Calculate its molecular formula.

How do you answer an empirical formula question?

What is the empirical formula for a compound containing 38.8 Carbon 16.2 hydrogen and 45.1 Nitrogen?

What are the 4 steps of calculating an empirical formula?

What is empirical formula with example?