Calculate empirical formula when given percent composition data
Problems 11 - 25

Ten Examples      Problems #1 - 10      Return to Mole Table of Contents

Determine empirical formula when given mass data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Determine molecular formula using the Ideal Gas Law

Determine the formula of a hydrate   Hydrate lab calculations


The example and problem questions only, no solutions


Problem #11: A compound is analyzed and found to consist of 34.5% Na, 23.3% P, and 42.1% O. What is its empirical formula?

Solution:

1) Assume 100 g of the compound is present. That means the percents can be expresse as mass:

Na ---> 34.5 g
P ---> 23.2 g
O ---> 42.1 g

2) Divide by gram-atomic weights:

Na ---> 34.5 g / 22.99 g/mol = 1.5
P ---> 23.2 g / 30.974 g/mol = 0.749
O ---> 42.1 g / 16.00 g/mol = 2.63125

3) Divide by smallest:

Na ---> 1.5 / 0.749 = 2
P ---> 0.749 / 0.749 = 1
O ---> 2.63125 / 0.749 = 3.5

4) Multiply through by 2 to get:

Na4P2O7

Problem #12: A compound is known to contain 4 moles of nitrogen per mole of compound. The elemental analysis shows 49.48% carbon, 28.85% nitrogen, 16.48% oxygen, and 5.19% of hydrogen. Determine its molecular formula.

Solution:

1) We need to know the empirical formula first. We start by assuming 100 g of the compound is present. This allows the percentages to be treated as masses. We then determine the moles of each element.

carbon ---> 49.48 g / 12.011 g/mol = 4.11956 mol
nitrogen ---> 28.85 g / 14.007 g/mol = 2.05968 mol
oxygen ---> 16.48 g / 16.00 g/mol = 1.03 mol
hydrogen ---> 5.19 g / 1.008 g/mol = 5.1488 mol

2) Divide through by the smallest value:

carbon ---> 4.11956 mol / 1.03 mol = 4
nitrogen ---> 2.05968 mol / 1.03 mol = 2
oxygen ---> 1.03 mol / 1.03 mol = 1
hydrogen ---> 5.1488 mol / 1.03 mol = 5

3) This is the empirical formula:

C4H5N2O

4) The empirical formula contains 2 moles of nitrogen for every mole of the formula. The molecular formula contains 4 moles per mole of formula. This is the molecular formula:

C8H10N4O2

Problem #13: A certain compound is composed of 55.81% carbon, 7.025% hydrogen and 37.17% oxygen with a molecular weight of 86.09 g/mol. What are the values of the subscripts x, y, and z if the molecular formula is CxHyOz?

Solution #1:

Assume 1.00 mole of the compound is present.

1) 55.81% of the compound is carbon. Therefore, 55.81% of the 86.09 is carbon:

86.09 g x 0.5581  
–––––––––––––––  =  4 <--- the value for x
12.01 g  

Note the use of 12.01 g to convert grams of C in 1 mole of the compound to moles of C in 1 mole of the compound.

2) 7.025% of the compound is hydrogen. Therefore, 7.025% of the 86.09 is hydrogen:

86.09 g x 0.07025  
––––––––––––––  =  6 <--- the value for y
1.008 g  

Note the use of 1.008 g to convert grams of H in 1 mole of the compound to moles of H in 1 mole of the compound.

3) 37.17% of the compound is oxygen. 55.81% of the 86.09 is oxygen

86.09 g x 0.3717  
––––––––––––––  =  2 <--- the value for z
16.00 g  

Note the use of 16.00 g to convert grams of O in 1 mole of the compound to moles of 0 in 1 mole of the compound.

4) The molecular formula:

C4H6O2

Solution #2:

Assume 100. g of the compound is present. This results in a slight longer solution.

1) Assuming 100 g of the compound turns the percentages into masses. Determine moles of each:

C ---> 55.81 g / 12.011 g/mol = 4.646574 mol
H ---> 7.025 g /1.008 g/mol = 6.969246 mol
O ---> 37.17 g / 16.00 g/mol = 2.323125 mol

2) Determine the smallest whole-number ratio:

C ---> 4.646574 mol / 2.323125 mol = 2
H ---> 6.969246 mol / 2.323125 mol = 3
O ---> 2.323125 mol / 2.323125 mol = 1

3) The empirical formula is:

C2H3O

4) Determine the molecular formula:

C2H3O weighs 43.0447.

86.09 / 43.0447 = 2

The molecular formula is C4H6O2


Problem #14: A compound contains 90.6% Pb and 9.4% O by weight. Determine the empirical formula.

Solution:

1) Assume 100 g of the compound is present. This means we have 90.6 g of Pb present and 9.4 g oxygen.

2) Determine moles of each:

Pb ---> 90.6 g / 207.2 g/mol = 0.43726 mol
O ---> 9.4 g / 16.00 g/mol = 0.5875 mol

3) Determine lowest whole-number ratio of moles:

Pb ---> 0.43726 mol / 0.43726 mol = 1
O ---> 0.5875 mol / 0.43726 mol = 1.3436

Note that 1.3436 is a fair representation of one and one-third (you can also think of it as four thirds).

Pb ---> 1 x 3 = 3
O ---> 1.3436 x 3 = 4.0308

4) The empirical formula is:

Pb3O4

Problem #15: A compound contains 70% Fe and 30% O by weight. Determine the empirical formula.

Solution:

1) Assume 100 g of the compound is present. This means we have 70 g of Fe present and 30 g oxygen.

2) Determine moles of each:

Fe ---> 70 g / 55.845 g/mol = 1.2535 mol
O ---> 30 g / 16.00 g/mol = 1.875 mol

3) Determine lowest whole-number ratio of moles:

Fe ---> 1.2535 mol / 1.2535 mol = 1
O ---> 1.875 mol / 1.2535 mol = 1.4958

Fe ---> 1 x 2 = 2
O ---> 1.4958 x 2 = 3

4) The empirical formula is:

Fe2O3

Problem #16: A compound that only contains carbon, hydrogen, and oxygen is 48.64% of C and 8.16% of H by mass. What is the empirical formula?

Solution:

1) Assume 100 g of the compound is present. This allows the percentages to be easily converted to masses:

C ---> 48.64 g
H ---> 8.16 g
O ---> 43.2 g

2) Convert the masses to moles:

C ---> 48.64 g / 12.011 g/mol = 4.0496 mol
H ---> 8.16 g / 1.008 g/mol = 8.095 mol
O ---> 43.2 g / 16.00 g/mol = 2.7 mol

3) Divide through by smallest:

C ---> 4.0496 mol / 2.7 mol = 1.5
H ---> 8.095 mol / 2.7 mol = 3
O ---> 2.7 mol / 2.7 mol = 1

4) Multiply 1.5, 3, 1 by 2 to obtain whole numbers (3, 6, 2).

empirical formula ---> C3H6O2

Problem #17: Determine the empirical formula of a compound that contains 44.099% C, 8.901% H, and 47.000% O.

Solution:

1) Assume that 100 g of the compound is present. This converts the percentages into masses. Determine moles:

C ---> 44.009 g / 12.011 g/mol = 3.664 mol
H ---> 8.901 g / 1.008 g/mol = 8.830 mol
O ---> 47.000 g / 16.00 g/mol = 2.9375 mol

2) Determine lowest whole-number ratio:

C ---> 3.664 mol / 2.9375 mol = 1.247
H ---> 8.830 mol / 2.9375 mol = 3
O ---> 2.9375 mol / 2.9375 mol = 1

3) The 1.247 associated with the carbon should not be rounded down. Instead, multiply by 4:

C ---> 1.247 x 4 = 4.988 = 5
H ---> 3 x 4 = 12
O ---> 1 x 4 = 4

The empirical formula is C5H12O4


Problem #18: Determine the empirical formula of a compound with the following mass percentages:

Carbon: 62%
Hydrogen: 10%
Oxygen: 28%

Solution:

Treat the percentages as actual values of a 100 g sample and then convert those gram values to mole values.

I end up with 5.1619 moles of carbon, 9.9212 moles of hydrogen, and 1.7501 moles of oxygen.

From there, I divide the moles of oxygen into the other values to try to find the whole-number mole proportions of each element.

C ---> 5.1619 / 1.7501 = 2.95
H ---> 9.9212 / 1.7501 = 5.67 <--- do NOT round to 6!!!!
O ---> 1.7501 / 1.7501 = 1

The 2.95 is close enough to round (to 3), but the 5.67 is a problem. So, I did this:

C ---> 9/3
H ---> 17/3
O ---> 3/3

Multiply by three to get the empirical formula:

C9H17O3

Problem #19: A oxide of phosphorus having a molar mass of 283.88 g/mol is 43.64% P by mass. What is the sum of the subscripts in the molecular formula?

Solution:

1) Let us assume 100 g of the compound is present. This allows the percents to be used as masses:

P ---> 43.64 g
O ---> 100 − 43.64 = 56.36 g

2) Let us convert to moles:

P ---> 43.64 g / 30.97 g/mol = 1.4091 mol
O ---> 56.36 g / 16.00 g/mol = 3.5225 mol

3) We want to know the smallest whole-number ratio between P and O:

P ---> 1.4091 mol / 1.4091 mol = 1
O ---> 3.5225 mol / 1.4091 mol = 2.5

4) Multiply by 2 to get a 2 and 5 and this is the empirical formula:

P2O5

5) What does one mole of the empirical formula weigh?

P2O5 ---> 141.943 g

6) We need to scale up the P2O5 to a formula that gives us the weight specified in the problem:

283.88 / 141.943 = 2

The 2 tells us that the P2O5 empirical formula must be doubled:

P4O10 <--- this is the molecular formula

The answer is 14, the sum of the subscripts of the formula that weighs 283.88 g/mol.


Problem #19: Carbohydrates are compounds of carbon, hydrogen, and oxygen in which the ratio of hydrogen atoms to oxygen atoms is 2:1. A certain carbohydrate is known to be 40% carbon. Its molar mass is 180 g/mol. Find the carbohydrate's empirical formula and molecular formula.

Solution:

1) Assume 180 g of the hydrocarbon is present.

(180 g) (0.40) = 72 g <--- that's the mass of carbon

2) Calculate moles of C in the one mole of the carbohydrate

72 g / 12 g/mol = 6

3) Calculate mass of the 2:1 H to O ratio

180 − 72 = 108 g

4) A 2 to 1 H to O ratio weighs 18 (One O is 16 and 2 H weighs 2)

108 / 18 = 6 <--- there are 6 H2O groups in the carbohydrate

5) 6 H2O groups is H12O6

C6H12O6

6) See how each subscript (the 6, 12, and 6) has a common factor of 6? Remove the common factor to get CH2O for the empirical formula.


Problem #20: A compound composed of only antimony and oxygen which is 83.53% Sb by mass and has a molar mass between 550 and 600 g/mol.

Solution:

1) Assume 100 g of the compound is present. This allows the percentages to be treated as masses. Convert the masses to moles.

Sb ---> 83.53 g / 121.760 g/mol = 0.68602 mol
O ---> 16.47 g / 16 g/mol = 1.029375 mol

2) Divide by smallest:

Sb ---> 0.68602 / 0.68602 = 1
O ---> 1.029375 / 0.68602 = 1.5

Multiply by 2 to get a whole-number ratio of 2 to 3.

3) Empirical and molecular formulas:

Empirical formula is Sb2O3

The weight of Sb2O3 is 291.517

(291.517) (2) = 583

The molecular formula is Sb4O6


Problem #21: A compound, made only from mercury and sulfur, is 86.2% Hg by mass. The molar mass of this compound is 465 g/mol. Determine the empirical formula and the molecular formula of this compound.

Solution:

86.2 g Hg ---> 0.43 moles Hg

and

13.8 g S ---> 0.43 moles S

so

Hg : S = 1 : 1

thus

(HgS)n = 465

where

n = 2

finally

Hg2S2

Comment: the ChemTeam did not write the above solution, but he does admit to being a bit jealous! :-)


Problem #22: A compound of carbon and sulphur has a composition of 15.8% carbon and 84.2% sulfur. What is the empirical formula?

Solution:

1) Consider 100 g of this compound

mass C = 15.8 g
mass S = 84.2 g

2) Determine moles of C and S:

moles C ---> 15.8 g/ 12.011 g/mol = 1.31546
moles S ---> 84.2 g/ 32.065 g/mol = 2.62592

3) Divide by the smallest number

C ---> 1.31546/ 1.31546 = 1
S ---> 2.62592 / 1.31546 = 1.996 = 2

4) CS2 is the empirical formula


Problem #23: The food additive E330 is an organic compound which occurs naturally in fruit. E330 has the following composition by mass: C, 37.5%; H, 4.17%; O, 58.3%. Calculate the empirical formula of E330.

Solution:

1) Assume 100 g of E330 is present. This converts percent amounts to grams.

2) Divide each mass by respective atomic mass:

C ---> 37.5 / 12.011 = 3.12214
H ---> 4.17 / 1.008 = 4.1369
O ---> 58.3 / 16.00 = 3.64375

3) Divide through by smallest

C ---> 3.12214 / 3.12214 = 1
H ---> 4.1369 / 3.12214 = 1.325
O ---> 3.64375 / 3.12214 = 1.167

4) Begin to remove fractions by multiplying by 3:

C = 3
H = 3.975
O = 3.5

5) 3.975 is pretty close to 4, but 3.5 requires another multiplying through, this time by 2

C = 6
H = 7.95 = 8
O = 7

6) Empirical formula = C6H8O7

7) Sometimes you see what the one factor required is (in this case, it was 6). But, sometimes, you also see your way through to the answer by using two factors. It other words, you might not have seen that six was the one step one, but you did see how a three works, followed up by a two.


Problem #24: An amino acid has a molar mass of 776.9 g/mole. It is 65.34% iodine by mass. How many iodine atoms are there per molecule of amino acid?

Solution:

1) Mass of iodine in one mole of the amino acid:

(776.9 g) (0.6534) = 507.62646 g

2) Moles of atoms of iodine present:

507.62646 g / 126.90447 g/mol = 4

3) There are 4 moles of iodine in every mole of the amino acid. There are 4 atoms of iodine in every molecule of the amino acid.


Problem #25: The elements X and Y form a compound that is 50% X and 50% Y by mass. The atomic mass of X is twice that of Y. What is the empirical formula of the compound?

Solution:

They each make up the same percentage by mass of the compound; so it will take twice as many Y atoms as X atoms to equal the same mass of Y compared to X.

XY2


Bonus Problem #1: Halothane is an anesthetic that is 12.17% C, 0.51% H, 40.48% Br, 17.96% Cl and 28.87% F by mass. What is the compound's molar mass if each molecule contains exactly one hydrogen atom? (Note: try and do this without a calculator.)

Solution:

Guess the formula as C2HBrClF3

How'd I do that?

Divide each percent by the atomic weight of the element and you get this:

C = 1
H = 0.5
Br = 0.5
Cl = 0.5
F = 1.5

Multiply through by 2.

I think the key #1 in this problem is to see that the 12.17% of carbon will go to 12.17 g and that 12.17 / 12.011 is essentially equal to 1. Key #2 is to see that hydrogen would be 0.51 g / 1.0 g/mol = 0.5 mole and that you would need to multiply it by 2 to get to one H atom. That means there will have to be two carbons.

The other elements are attacked in the same way.


Bonus Problem #2: What is the empirical formula if you have 81.71% carbon and 18.29% hydrogen?

Solution:

1) Assume 100 g of the compound is present. Based on the percentages, that means 81.81g of carbon and 18.18 g of hydrogen are present. Calculate moles of each:

carbon ---> 81.71g / 12.011 g/mol = 6.80293 mol
hydrogen ---> 18.29 g / 1.008 g/mol = 18.14484 mol

2) Divide by smallest:

carbon ---> 6.80293 mol / 6.80293 mol = 1
hydrogen ---> 18.14484 mol / 6.80293 mol = 2.6672

3) Do not round 2.6672 to 3. It is too far away. Instead, see that it is two and two-thirds. The proper choice is to multiply by 3:

carbon ---> 1 x 3 = 3
hydrogen ---> 2.6672 x 3 = 8

The empirical formula is C3H8.


Ten Examples      Problems #1 - 10      Return to Mole Table of Contents

Determine empirical formula when given mass data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Determine molecular formula using the Ideal Gas Law

Determine the formula of a hydrate   Hydrate lab calculations