SCH 3U Moles Name: _________________________
Empirical Formula Calculations
The empirical formula (simplest formula) shows the simplest whole number ratio of the atoms of each element
in a compound. This may not be the actual or true formula because the number of atoms in the actual formula may have
been reduced to the simplest ratio (eg. C2H4 has an empirical formula of CH2).
The calculation of empirical formula is essentially the reverse process of calculating percent composition. The
cart below outlines the steps to calculating empirical or simplest formula.
species mass or %
divide by smallest from column 3
reduce to smallest whole number ratio
Examples:
1. Analysis of a compound has determined that it contains 29.08 % Na, 40.58 % S and 30.34 % O. Find the empirical
formula of this compound.
Solution:
species mass or %
divide by smallest from column 3
reduce to smallest whole number ratio
Na 29.08
2
S 40.58
2
O 30.34
3
Therefore the empirical or simplest formula is Na2S2O3.
2. A 5.72 g sample of washing soda (hydrated sodium carbonate, Na2CO3 · x H2O) was heated to give 2.12 g of
anhydrous sodium carbonate (Na2CO3). What is the simplest formula of the washing soda?
Solution:
species mass or %
divide by smallest from column 3
reduce to smallest whole number ratio
Na2CO3 2.12
1
H2O 3.60
10
Therefore the simplest formula is Na2CO3 · 10 H2O.
SCH 3U Moles Name: _________________________
Exercise:
1. Methane gas was determined to be 75 % C and 25 %H by mass. Determine its empirical formula.
2. Calculate the simplest formula for the compounds whose compositions are listed below.
a) 15.75 % C, 84.25 % S
b) 31.13 % S, 68.87 % Cl
c) 81.61 % Cl, 18.39 % O
d) 70.11 % Ag, 9.10 % N, 20.79 % O
e) 40.27 % K, 26.78 % Cr, 32.95 % O
f) 26.58 % K, 35.35 % Cr, 38.07 % O
g) 27.38 % Na, 1.19 % H, 14.29 % C, 57.14 % O
h) 88.52 % B, 11.48 % H
3. An analysis of nicotine gave 74.07% C, 8.64 % H, 17.28 % N. What is the simplest formula for nicotine?
4. A hydrated compound was found to contain 13.92 % Ca, 1.39 % H, 21.52 % P, 44.43 % O and 18.74 % water.
What is the simplest formula of the compound?
5. A 7.81 g sample of an unknown compound was broken down to give 4.60 g of sodium and 3.21 g of sulfur.
What is the simplest formula for this compound?
6. When the oxides of mercury are heated, they decompose to mercury and oxygen gas. If a 1.083 g sample of
an oxide of mercury produces 1.003 g, what is the empirical formula of this compound?
7. A student wishes to find the formula of a compound formed when copper is heated is heated in the
presence of molten sulfur. Use the data below collected by the student to determine the simplest formula.
mass of empty crucible and lid 24.60 g
mass of crucible, lid and copper 25.87 g
mass of crucible, lid and product 26.19 g
8. It was determined that a 10.66 g sample of a pure substance contained 3.91 g of potassium, 3.55 g of
chlorine and 3.20 g of oxygen. What is the empirical formula of this compound?
9. The element “M” forms the chloride MCl4. If the chloride contains 75 % chlorine by mass, calculate the
molar mass of element “M”.
10. The elements “X” and “Y” combine to form a compound which is 40 % “X” by mass. If the molar mass of “X”
is twice that of “Y”, what is the simplest formula of the compound formed?
Answers:
1. CH4
2. a) CS2
b) SCl2
c) Cl2O
d) AgNO2
e) K2CrO4
f) K2Cr2O7
g) NaHCO3
h) B5H7
3. C5H7N
4. CaH4P2O8 · 3 H2O or Ca(H2PO4)2 · 3 H2O
5. Na2S
6. HgO
7. Cu2S
8. KClO3
9. 47.3 g/mol
10. XY3
