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Wize University Chemistry Textbook > Stoichiometry

Empirical and Molecular Formulae Problems

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Example: Empirical and Molecular Formula (Problem Type #1)

(Finding the Empirical and Molecular Formula Given Percent Compositions)

Wize Tip
To find the empirical and molecular formulae given percent compositions, follow these steps:
Step 1 – Find the mass % composition of each compound
Step 2 – Assume 100 g sample and calculate the grams of each element
Step 3 – Calculate the moles of each element
Step 4 – Divide each number of moles by the greatest common factor
Step 5 – Write out the empirical formula!
Step 6 – If given the molecular weight of the molecule you can now find the molecular formula
  • molar mass of empirical formula(x) = molar mass of molecule
  • Multiply each subscript in the empirical formula by x to get the molecular formula!


PAGE BREAK
Analysis of an unknown compound tells us that a sample has 39% C, 10% H, and 51% Oxygen. The molar mass of the compound was found to be 124g/mol. What is the empirical and molecular formula for this compound?

Step 1 – Find the mass % composition of each compound

Here we already have been given the percent composition by mass of each element:

C makes up
39
%, H makes up
10
%, and O makes up
51
% of the sample by mass

Step 2 – Assume 100 g sample and calculate the grams of each element

If we assume a 100g sample, how many grams would be accounted for by C, H, and O?

C=
39
g H=
10
g O=
51
g

Step 3 – Calculate the moles of each element n=mMn=\frac{m}{M}


C: n=3912n=\frac{39}{12} H: n=101n=\frac{10}{1} O: n=5116n=\frac{51}{16}

n∼3 moles n\sim3\ moles\ n=10 moles n=10\ moles\ n∼3 molesn\sim3\ moles

Step 4 – Divide each number of moles by the greatest common factor

C: 33=1\frac{3}{3}=1 H: 103∼3\frac{10}{3}\sim3 O: 33=1\frac{3}{3}=1

Step 5 – Write out the empirical formula!

Now we know that for every
1
C, there are
3
H and
1
O

Based on this information, we can write the empirical formula:
CH3O

Step 6 – If given the molecular weight of the molecule you can now find the molecular formula

  • We are told the molar mass of the whole molecule is: 124g/mol
  • If we find the molar mass of the empirical formula, we can see if we need to multiply it by a certain number to get the molar mass of the overall compound:
M(CH3O)=12g/mol + 3(1g/mol) + 16g/mol
M(CH3O)=31g/mol

Total mass of the molecule is 124 g/mol
(124g/mol)/31g/mol=4

This means we must multiply the subscripts in the empirical formula by 4 to get the molecular formula:
CH3O -----x4----> C4H12O4

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Example: Empirical and Molecular Formula (Problem Type #2)

(Finding the Empirical and Molecular Formulae Given Masses in Grams of Elements)


A sample of a compound contains 1.52g of N atoms and 3.47g of O atoms. The molar mass of the compound is between 90.0g and 95.0g. Determine the empirical and molecular formulas. Also, calculate the actual molar mass of this compound.


We have the masses of each element. Convert those masses to moles and use molar ratio to find the empirical formula.

1) Calculate the moles of each element
# mol N=1.52g N×mol14.01g=0.108 mol N\#\ \text{mol N}=1.52 \text{g N}\times\dfrac{\text{mol}}{14.01\text{g}}=0.108 \text{ mol N}

# mol O=3.47 g Omol15.9994 g=0.217 mol O\#\text{ mol O}=3.47 \text{ g O}\dfrac{\text{mol}}{15.9994\ \text{g}}=0.217 \text{ mol O}


2) Now divide each value by the smallest number of mol to find the empirical formula…

N0.1080.108O0.2170.108∴N1O2orNO2\dfrac{\text{N}_{0.108}}{0.108}\dfrac{\text{O}_{0.217}}{0.108}\quad\therefore\quad\text{N}_1\text{O}_2\quad or \quad\text{NO}_2




3) Determine the molecular formula
  • The empirical formula has a molar mass of 46.01 g/mol.
  • We know the range of the molecular formula mass, so select a mass in that range and find out the multiplicative coefficient for the molecular formula.

Molecular MassEmpirical Mass=90 g/mol46 g/mol=∼2\dfrac{\text{Molecular Mass}}{\text{Empirical Mass}}=\dfrac{90\text{ g/mol}}{46\text{ g/mol}}= \sim2

∴N2O4\therefore \quad \text{N}_2\text{O}_4 is the molecular formula!

Practice: Molecular Formula from Empirical Formula

p-dichlorobenzene (mM = 147.00 g/mol) is a common disinfectant and deodorant used in industrial applications. It has an empirical formula of C3H2Cl. Determine its molecular formula.

Note: if your answer is C3H2Cl, enter it as: C3H2Cl.


Practice: Determine Empirical and Molecular Formulae

A compound is found to contain 50.05% sulfur and 49.95% oxygen by weight. The molecular weight for this compound is 64.07 g/mol.
What is the empirical formula for this compound?
Extra Practice
Mass Percentage
A compound containing sodium, chlorine, and oxygen is 25.42 % sodium by mass. A 3.25 g sample gives 4.33 x 1022 atoms of oxygen. What is the empirical formula?
MW Na: 23 g/mol
MW Cl: 35.45 g/mol
Empirical formula
A compound has the following mass composition: C 86.59%, H 8.36% and N 5.05%. What is the empirical formula of this compound?
Gas Stoichiometry
Cyanogen is 46.2 % C and 53.8 % N by mass. 1.00 g cyanogen gas occupies 0.476 L at 25 °C and 750 Torr. Determine the empirical and molecular formula of cyanogen.
Mass Percentage II
A certain metal hydroxide with the formula M(OH)2 is 17.22 % oxygen by mass. What is the identity of the metal, "M" ?
Spring 2017, Practice Exam 1 (CHM2045/UF)
A 7.21 g sample of a LiClO4 · xH2O was heated to drive off the water. The mass was reduced to 4.78 g. What is the formula of the hydrate?
Mass Percentage
A compound containing sodium, chlorine, and oxygen is 25.42 % sodium by mass. A 3.25 g sample gives 4.33 x 1022 atoms of oxygen. What is the empirical formula?
Hint: This is a tricky problem but start by writing the variables that you know.....what can you solve for?
If you're still not sure, my hint is to start with O! We can calculate moles of O, then mass of O, then % mass of O! What should we do next??
Practice: Empirical and Molecular Formula
A sample of a compound contains 1.52 g of N atoms and 3.47 g of O atoms. The molar mass of the compound is between 90.0 g and 95.0 g. Determine the empirical and molecular formulas.
Elements: Mass Percentage
A metal oxide with the formula M2O3 is 21.2 % oxygen by mass. What is the identity of the metal, "M"?
Empirical formula: Hydrates
A 15.67 g sample of a hydrate of magnesium carbonate (MgCO3) was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?
Empirical Formulae
A compound has an empirical formula of C2H5O. Its molar mass is 225.305 g/mol. Determine its molecular formula.
Empirical Formulae
7. A compound has an empirical formula of CH2O. Its molar mass is 180.01 g/mol. Determine its molecular formula.
Chemical formula: Molecular formula
Part 1: What's the empirical formula for a molecule containing 65.5% C, 5.5% H, and 29.0% O?
Part 2: It was found that the atomic mass of the molecule is 110g/mol. What is the molecular formula?
Molecular Formula from Empirical Formula
p-dichlorobenzene (mM = 147.00 g/mol) is a common disinfectant and deodorant used in industrial applications. It has an empirical formula of C3H2Cl. Determine its molecular formula. You do not need subscripts for your submission. For example, if you answer is the empirical formula, enter it as: C3H2Cl.
Chemical Reactions: Finding an Unknown Empirical Formula from Mass Percentages
A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33 x 1022 atoms of oxygen. What is the empirical formula?
Chemical Reactions: Hydrate Question
A 15.67 g sample of a hydrate of magnesium carbonate (MgCO3) was heated, without decomposing the carbonate, to drive off the water. The mass was reduced to 7.58 g. What is the formula of the hydrate?
Chemical Formula: Mass Percentage Problem
A compound is found to contain 50.05 % sulfur and 49.95 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 64.07 g/mol. What is its molecular formula?
Note: Empirical formula is the formula that expresses the SMALLEST whole number ratio of atoms present whereas the molecular formula expresses the same ratio as the empirical formula but is the actual number of atoms present in a molecule.
Practice: Mass Percentage
A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives 4.33 x 1022 atoms of oxygen. What is the empirical formula?
Hint: This is a tricky problem but start by writing the variables that you know.....what can you solve for?
If you're still not sure, my hint is to start with O! We can calculate moles of O, then mass of O, then % mass of O! What should we do next??
Empirical and Molecular Formula: Molar Mass
A sample of a compound contains 1.52 g of N atoms and 3.47 g of O atoms. The molar mass of the compound is between 90.0 g and 95.0 g. Determine the empirical and molecular formulas. Also, calculate the actual molar mass of this compound.