How do you calculate the empirical formula?

The simplest formula or the empirical formula provides the lowest whole number ratio of atoms existent in a compound. The relative number of atoms of every element in the compound is provided by this formula.

Steps for Determining an Empirical Formula

• Let’s begin with what’s given in the problem, i.e., the number of grams of each element.
• We’ll assume that the total mass is 100 grams if percentages are given, so that

Each element’s mass = the given percentage

• By making use of the molar mass from the periodic table, change the mass of every element to moles.
• Divide every mole value by the lowest number of moles computed.
• Round up to the closest whole number.  This is denoted by subscripts in the empirical formula and is the mole ratio of the elements.

Multiply each answer by the same factor to get the lowest whole number multiple, if the number is too far to round off (x.1 ~ x.9).

e.g.  Multiply each solution in the problem by 4 to get 5, if one solution is 1.25.

e.g.  Multiply each solution in the problem by 2 to get 3, if one solution is 1.5.

The molecular formula can be calculated for a compound if the molar mass of the compound is given when the empirical formula is found.  To find the ratio between the molecular formula and the empirical formula. Basically, the mass of the empirical formula can be computed by dividing the molar mass of the compound by it.  Multiply every atom (subscripts) by this ratio to compute the molecular formula.

Solved Examples

Problem 1: A compound contains 88.79% oxygen (O) and 11.19% hydrogen (H). Compute the empirical formula of the compound.

Solution:

1. Assume 100.0g of substance. We see that the percentage of each element matches the grams of each element

11.19g H

88.79g O

1. Convert grams of each element to moles

H: (11.19/1.008) = 11.10 mol H atoms [molar mass of H=1.008g/mol]

O: (88.79/16.00) = 5.549 mol O atoms [molar mass of O= 16.00g/mol]

The formula could be articulated as H11.10O5.549. However, it’s usual to use the smallest whole number ratio of         atoms

1. By dividing the lowest number alter the numbers to whole numbers.

H =11.10/ 5.549  = 2.000

O = 5.549/ 5.549= 1.000

The simplest ratio of H to O is 2:1

Empirical formula = H2O

Problem 2: A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?

If carbon and hydrogen are present in a compound in a ratio of 1:2, the empirical formula for the compound is CH2.

The molecular formula for the same compound will equal to n × (CH2), in other words, the molecular formula for this compound will be CnH2n

Using the periodic table of the elements we can determine the empirical formula mass of this compound(CH2): 12.01 + (2 × 1.008) = 14.026

If we know the molecular mass of the compound is 28.00 (Mr = 28.00) then we can find the value of "n" in the molecular formula CnH2n:

Write the general expression:Mr=n × empirical formula massSubstitute in the known values:28.00=n × 14.026Divide both sides of equation by 14.02628.00
14.026=n × 14.026
14.026Solve for n2=n

Subsitute this value, n = 2, back into the general molecular formula CnH2n to get the molecular formula of this compound.
So the molecular formula for the compound is:

C(1 × 2)H(2 × 2) which is C2H4

There are many compounds that can have the empirical formula CH2 and therefore a molecular formula of the form CnH2n.

Examples include:

• C2H4 (ethene or ethylene) molecular mass=28.0 and n = 2, that is, C(1 × 2)H(2 × 2)
• C3H6 (propene or propylene) molecular mass=42.0 and n = 3, that is, C(1 × 3)H(2 × 3)
• C3H6 (cyclopropane) molecular mass=42.0 and n = 3, that is, C(1 × 3)H(2 × 3)
• C4H8 (butene or butylene(2)) molecular mass=56.0 and n = 4, that is, C(1 × 4)H(2 × 4)
• C4H8 (cyclobutane) molecular mass=56.0 and n = 4, that is, C(1 × 4)H(2 × 4)

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Theory Behind Calculating Empirical Formula from Percentage Composition

We can use the percentage composition (percent composition) of a compound to determine an empirical formula for the compound.

Remember that the percentage composition gives us the percent by mass of each element present in the compound, for example

If a compound is made up of 10% by mass of element X and 90% by mass of element Z
Then 100 g of the compound will contain:

• 10 g of element X

  mass of X =10 %×100 g=10
  100× 100= 10 g

• 90 g of element Z

  mass of Z =90 %×100 g=90
  100× 100= 90 g

10 g of element X and 90 g of element Z is present in 100 g of the compound.

On the other hand, the empirical formula of the compound gives us the lowest whole number ratio of atoms of each element in the compound, for example

Empirical formula XZ tells us that for every 1 atom of X there is 1 atom of Z in the compound.

Empirical formula XZ2 tells us that for every 1 atom of X there are 2 atoms of Z, etc

Clearly, we will need to convert the "mass" of each element in the compound into a "number of atoms".

6.02 × 1023 atoms of element X = 1 mole of atoms of element X

Mass of 1 mole of atoms of element X = molecular mass expressed in grams (molar mass)

So we can use the "mass" and "molar mass" (or relative molecular mass) to calculate the moles of atoms of each element present,

moles of element X =        mass of element X in grams        
molar mass of element X in grams per molemoles of element Z =        mass of element Z in grams        
molar mass of element Z in grams per mole

Next we need to write the relationship between moles of X and moles of Z in the compound:

moles of element X:moles of element Z

and then to find the lowest whole number ratio of these moles

• If moles of Z < moles of X, divide both by moles of Z

  moles of element X
  moles of element Z:moles of element Z
  moles of element Zthat ismoles of element X
  moles of element Z: 1

• If moles of X < moles of Z, divide both by moles of X

  moles of element X
  moles of element X:moles of element Z
  moles of element Xthat is1 :moles of element Z
  moles of element X

Be aware that the calculation above to get the mole ratio of one element to another may not result in whole numbers.
A further calculation may be needed to get the lowest whole number ratio of moles of one element to another.
This is where a recognition of the decimal equivalent of common fractions, as shown in the table below, can be very helpful:

Common DecimalEquivalent FractionMole Ratio Example0.1251/81 : 1.125 converts to 1 : 9/8
multiply throughout by 8 to give 8 : 90.251/41 : 0.25 converts to 1 : 1/4
multiply throughout by 4 to give 4 : 10.331/31 : 1.33 converts to 1 : 4/3
multiple throughout by 3 to give 3 : 40.3753/81 : 1.375 converts to 1 : 11/8
multiply throughout by 8 to give 8 : 110.51/22 : 1.5 converts to 2 : 3/2
multiply throughout by 2 to give 4 : 30.6255/81 : 1.625 converts to 1 : 13/8
multiply throughout by 8 to give 8 : 130.6672/32 : 1.66 converts to 2 : 5/3
multiply throughout by 3 to give 6 : 50.8757/81 : 0.875 converts to 1 : 7/8
multiply throughout by 8 to give 8 : 7

Once you have calculated the lowest whole number ratio of moles of one element to the other, you are ready to write the empirical formula, for example

• If the mole ratio of X to Z is 1:1, empirical formula is XZ
• If mole ratio of X to Z is 1:2, empirical formula is XZ2
• If mole ratio of X to Z is 2:3, empirical formula is X2Z3

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Steps for Calculating Empirical Formula from Percentage Composition

1. Assume 100 g of sample
2. Convert all percentages to a mass in grams
3. Find the relative atomic mass of each element present using the Periodic Table
4. Calculate the moles of each element present: n = mass ÷ relative atomic mass
5. Divide the moles of each element by the smallest of these to get a mole ratio
6. If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula
7. If the numbers in the mole ratio are NOT whole numbers, you will need to further manipulate these until the mole ratio is a ratio of whole numbers (integers)

Worked Example: Empirical Formula from Percentage Composition

Question: A compound is found to contain 47.25% copper and 52.75% chlorine. Find the empirical formula for this compound.

Solution:

1. Assume mass of sample of compound is 100 g
2. Convert all percentages to a mass in grams

    ElementsCuCl% by mass47.2552.75(b) mass in 100 g of compound47.25
   100× 100= 47.25 g52.75
   100× 100= 52.75 g

3. Find the relative atomic mass (molar mass) of each element present using the Periodic Table

    ElementsCuCl% by mass47.2552.75(b) mass in 100 g of compound47.25
   100× 100= 47.25 g52.75
   100× 100= 52.75 g(c) molar mass / g mol-163.5535.45

4. Calculate the moles of each element present: n = mass ÷ relative atomic mass

    ElementsCuCl% by mass47.2552.75(b) mass in 100 g of compound47.25
   100× 100= 47.25 g52.75
   100× 100= 52.75 g(c) molar mass / g mol-163.5535.45(d) moles / mol47.25
   63.55= 0.743552.75
   35.45= 1.488

5. Divide the moles of each element by the smallest of these to get a mole ratio

    ElementsCuCl% by mass47.2552.75(b) mass in 100 g of compound47.25
   100× 100= 47.25 g52.75
   100× 100= 52.75 g(c) molar mass / g mol-163.5535.45(d) moles / mol47.25
   63.55= 0.743552.75
   35.45= 1.488(e) divide moles by 0.74350.7435
   0.7435= 11.488
   0.7435= 2

6. If the numbers in the mole ratio are all whole numbers (integers) convert this to an empirical formula:

   mole ratio Cu : Cl is1:2 (1 and 2 are both whole numbers)empirical formula isCuCl2

7. If the numbers in the mole ratio are NOT whole numbers, you will need to further manipulate these until the mole ratio is a ratio of whole numbers (integers)
   This step is not required since we have already achieved the lowest ratio of whole numbers

Empirical formula for this compound is CuCl2

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Worked Example: Molecular Formula from Percentage Composition and Molar Mass

Question: A compound with a molar mass of 34.0 g mol-1 is known to contain 5.88% hydrogen and 94.12% oxygen. Find the molecular formula for this compound.

How do you find the empirical formula?

How do you find N in empirical formula?

Why do we calculate empirical formula?

How do you calculate the empirical formula mass?