### Calculate empirical formula when given percent composition dataTen Examples

Notice below how I do the first problem with some attention to using proper atomic weights, as well as keeping close to the proper number of significant figures. Then, notice how I get away from that (as well as being real consistent with units) in the following problems.

Notice also how it really doesn't make much of a difference. The trick is to know when to do that and it comes only via experience. Generally speaking, in empirical formula problems, C = 12, H = 1, O = 16 and S = 32 are sufficient.

There are times when using 12.011 or 1.008 will be necessary. If you hit a problem that just doesn't seem to be working out, go back and re-calculate with more precise atomic weights. These problems, however, are fairly uncommon.

For what it is worth, one piece of advice on rounding: don't round off on the moles if you see something like 2.33 or 4.665. That first one can be rendered as two and one-third (or seven thirds) and the second one as four and two-thirds (or fourteen thirds). In a situation like that, you would multiply by three to reach the smallest whole-number ratio rather than dividing by the smallest.

I know it's easy to say, harder to demonstrate. Some of the problems below involve this thirds issue. Look for a problem involving citric acid. Just be aware that rounding off too early and/or too much is a common problem in this type of problem.

Example #1: 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?

Solution:

1) Assume 100 g of the compound is present. This changes the percents to grams:

S ---> 50.05 g
O ---> 49.95 g

2) Convert the masses to moles:

S ---> 50.05 g / 32.066 g/mol = 1.5608 mol
O ---> 49.95 g / 16.00 g/mol = 3.1212 mol

3) Divide by the lowest, seeking the smallest whole-number ratio:

S ---> 1.5608 / 1.5608 = 1
O ---> 3.1212 / 1.5608 = 2

4) Write the empirical formula:

SO2

5) Compute the "empirical formula weight" (I'll use EFW from now on):

32 + 16 + 16 = 64

6) Divide the molecule weight by the "EFW:"

64.07 / 64 = 1

7) Use the scaling factor computed just above to determine the molecular formula:

SO2 times 1 gives SO2 for the molecular formula

Example #2: A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for this compound? The molecular weight for this compound is 74.14 g/mol. What is its molecular formula?

Solution:

1) Assume 100 g of the compound is present. This changes the percents to grams:

C ---> 64.80 g
H ---> 13.62 g
O ---> 21.58 g

2) Convert the masses to moles:

C ---> 64.80 g / 12 = 5.4
H ---> 13.62 g / 1 = 13.62
O ---> 21.58 g / 16 = 1.349

3) Divide by the lowest, seeking the smallest whole-number ratio:

C ---> 5.4 / 1.349 = 4
H ---> 13.62 / 1.349 = 10
O ---> 1.349 / 1.349 = 1

4) Write the empirical formula:

C4H10O

5) Determine the molecular formula:

"EFW" ---> 48 + 10 + 16 = 74

74.14 / 74 = 1

molecular formula = C4H10O

Example #3: A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. What is the empirical formula for this compound? The molecular weight for this compound is 102.2 g/mol. What is its molecular formula?

Solution:

1) Percents to mass, based on assuming 100 g of compound present:

S ---> 31.42 g
O ---> 31.35 g
F ---> 37.23 g

2) Calculate moles of each:

S ---> 0.982 mol
O ---> 1.96 mol
F ---> 1.96 mol

3) Smallest whole-number ratio:

S ---> 1
O ---> 2
F ---> 2

4) Write the empirical and molecular formula formula:

SO2F2

"EFW" ---> 32 + 32 + 38 = 102 g

the empirical formula is also the molecular formula

Example #4: Ammonia reacts with phosphoric acid to form a compound that contains 28.2% nitrogen, 20.8% phosphorous, 8.1% hydrogen and 42.9% oxygen. Calculate the empirical formula of this compound.

Solution:

1) Masses:

N ---> 28.2 g
P ---> 20.8 g
O ---> 42.9 g
H ---> 8.1 g

2) Moles:

N ---> 2
P ---> 0.67
O ---> 2.68
H ---> 8

3) Lowest whole-number ratio:

N = 2 / 0.67 = 3
P = 0.67 / 0.67 = 1
O = 2.68 / 0.67 = 4
H = 8 / 0.67 = 12

4) Empirical formula:

N3H12PO4

or

(NH4)3PO4

Although not asked for, the name of this compound is ammonium phosphate.

5) I would like to discuss my piece of advice (about thirds) at the top of the file using the moles data from the above problem.

N ---> 2 = 6/3
P ---> 0.67 = 2/3
O ---> 2.68 = 8/3
H ---> 8 = 24/3

6) Then, I multiply:

N ---> 6/3 times 3 = 6
P ---> 2/3 times 3 = 2
O ---> 8/3 times 3 = 8
H ---> 24/3 times 3 = 24

7) Notice how doing it this way introduces an extra factor of 2. We remove the extra factor of two to arrive at this ratio:

N ---> 3
P ---> 1
O ---> 4
H ---> 12

8) The extra factor of two could have also been removed like this:

N ---> 3/3
P ---> 1/3
O ---> 4/3
H ---> 12/3

And then a multiply through by 3 yields the 3, 1, 4, 12 mentioned in step 7

8) And we continue on. I really don't want you to think that the introduction of the extra factor of two damages this technique. There are times when changing everything to third-type fractions will make things easier. As in the following example.

Example #5: A compound contains 57.54% C, 3.45% H, and 39.01% F. What is its empirical formula?

Solution:

1) Get mass, then moles:

C ---> 57.54 / 12.011 = 4.791
H ---> 3.45 / 1.0008 = 3.423
F ---> 39.01 / 19.00 = 2.053

2) Seek lowest whole-number ratio:

C ---> 4.791 / 2.053 = 2.33
H ---> 3.423 / 2.053 = 1.67
F ---> 2.053 / 2.053 = 1

3) The key here is to see that 2.33 is 2 and one-third or 7/3 and that 1.67 is 5/3. Therefore:

C ---> (7/3) x 3 = 7
H ---> (5/3) x 3 = 5
F ---> (3/3) x 3 = 3

Empirical formula is C7H5F3

Example #6: Vanillin, the flavoring agent in vanilla, has a mass percent composition of 63.15%C, 5.30%H, and 31.55%O. Determine the empirical formula of vanillin.

Solution:

1) Start by assuming 100 g is present, therefore:

C ---> 63.15 g
H ---> 5.30 g
O ---> 31.55 g

2) Determine moles:

C ---> 63.15 g / 12.011 g/mol = 5.25768 mol
H ---> 5.30 g / 1.008 g/mol = 5.25794 mol
O ---> 31.55 g / 16.00 g/mol = 1.97188 mol

3) Divide through by lowest value:

C ---> 5.25768 mol / 1.97188 mol = 2.67
H ---> 5.25794 mol / 1.97188 mol = 2.67
O ---> 1.97188 mol / 1.97188 mol = 1

4) Do not round off the 2.67 to 3. Think of 2.67 as 2 and two-thirds, which becomes 8/3. Multiply the above through by 3 to get this:

C ---> 2.67 x 3 = 8
H ---> 2.67 x 3 = 8
O ---> 1 x 3 = 3

5) Empirical formula is C8H8O3, not the C3H3O you would get by rounding 2.67 to 3.

Example #7: A compound was found to contain 24.74% (by mass) potassium, 34.76% manganese, and 40.50% oxygen. Determine the empirical formula.

Solution:

1) Collect atomic mass:

Potassium (K) has 39.1 amu
Mnaganese (Mn) has 54.9 amu
Oxygen (O) has 16.0 amu

2) Calculate stoichiometric ratio:

K ---> 24.74 / 39.1 = 0.63
Mn ---> 34.76 / 54.9 = 0.63
O ---> 40.50 / 16.0 = 2.53

3) Find integer numbers on the basis of ratios:

K : Mn : O = 0.63 : 0.63 : 2.53 = 1 : 1 : 4

4) Write empirical formula:

KMnO4

Example #8: A mass spectrometer analysis finds that a molecule has a composition of 48% Cd, 20.8% C, 2.62% H, 27.8% O. Determine the empirical formula.

Solution:

1) Let us assume 100 g of the compound is present. This means:

48 g Cd, 20.8 g C, 2.62 g H, 27.8 g O

2) Let us determine moles present:

Cd ---> 48 g / 112.4 g/mol = 0.427 mol
C ---> 20.8 g / 12.011 g/mol = 1.732 mol
H ---> 2.62 g / 1.008 g/mol = 2.5992 mol
O ---> 27.8 g / 16.00 g/mol = 1.7375 mol

3) Divide through by lowest value:

Cd ---> 0.427 mol / 0.427 mol = 1
C ---> 1.732 mol / 0.427 mol = 4.06
H ---> 2.5992 mol / 0.427 mol = 6.09
O ---> 1.7375 mol / 0.427 mol = 4.07

4) Ignore the Cd and see a 4 : 6 : 4 ratio for C : H : O. Reduce it to 2 : 3 : 2. Therefore:

C2H3O2

C2H3O2¯ is the acetate ion

5) Cadmium is divalent, so we can see the empirical formula as:

Cd(C2H3O2)2

Notice how the molar ratio in the full formula for cadium acetate is 1 : 4 : 6 : 4

Example #9: A bromoalkane contains 35% carbon and 6.57% hydrogen by mass. Calculate the empirical formula of this bromoalkane.

Solution:

1) Assume 100 g of the compound is available:

C ---> 35 g
H ---> 6.57 g
Br ---> 58.43 g (from 100 − 41.57)

2) Determine moles:

C ---> 35 g / 12 gmol = 2.917
H ---> 6.57 g / 1 g/mol = 6.57
Br ---> 58.43 g / 80 g/mol = 0.730375

3) Divide by smallest to seek lowest whole-number ratio:

C ---> 2.917 / 0.730375 = 4
H ---> 6.57 / 0.730375 = 9
Br ---> 0.730375 / 0.730375 = 1

C4H9Br

Example #10: 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?

Solution:

1) Percent oxygen in the sample:

4.33 x 1022 atoms / 6.022 x 1023 atoms/mol = 0.071903 mol

(0.071903 mol) (16.00 g/mol) = 1.15045 g

1.15045 g / 3.25 g = 0.3540 = 35.40%

2) Percent chlorine:

100 − (25.42 + 35.40) = 39.18%

3) Assume 100 g of the compound is present. This converts percents to grams. Determine moles:

Na ---> 25.42 g / 23.0 g/mol = 1.105
Cl ---> 39.18 g / 35.453 g/mol = 1.105
O ---> 35.40 g / 16.00 g/mol = 2.2125

4) Finish with lowest whole-number ratio:

Divide by 1.105 to get lowest whole-number ratio of 1 : 1 : 2

NaClO2

Although not asked for, this is the formula for sodium chlorite.

Bonus Example: A chemist observed a gas being evolved in a chemical reaction and collected some of it for analyses. It was found to contain 80% carbon and 20% hydrogen. It was also observed that 500. mL of the gas at STP weighed 0.6695 g. What is the empirical formula for the compound? What is its molecular formula?

Solution:

1) Determine the empirical formula:

Assume 100 g of the compound is present.

That means 80 g of C and 20 g of H.

That means 6.67 mole of C and 20 mole of H.

The above molar ratio is 1:3, meaning the empirical formula is CH3

2) Determine the molar mass of the compound:

Since everything is at STP, I can use molar volume.

 22.414 L 0.500 L ––––––– = ––––––– 1.00 mol x

x = 0.0223075 mol

molar mass ---> 0.6695 g / 0.0223075 mol = 30.0 g/mol

3) Determine the molecular formula:

The "empirical formula weight" (not a standard term in chemistry) of CH3 is 15.

30 / 15 = 2

Molecular formula is C2H6.