### Calculate empirical formula when given percent composition dataProblems 1 - 10

Problem #1: Analysis of a compound containing only C and Br revealed that it contains 33.33% C atoms by number and has a molar mass of 515.46 g/mol. What is the molecular formula of this compound?

Solution:

1) ". . . 33.33% C atoms by number . . ." Since mole is a measure of how many (one mole = 6.022 x 1023 chemical entities), we know this:

C ---> 0.3333 mol
Br ---> 0.6667 mol

2) Let us determine the smallest whole-number ratio:

C ---> 0.3333 / 0.3333 = 1
Br ---> 0.6667 / 0.3333 = 2

3) The empirical formula is CBr2. Determine the molecular formula:

515.46 / 171.819 = 3

C3Br6

Problem #2: Chemical analysis shows that citric acid contains 37.51% C, 4.20% H, and 58.29% O. What is the empirical formula?

Solution:

1) We start by assuming 100 g of the compound is present. This turns the above percents into masses.

2) Calculate moles:

C ---> 37.51 / 12.011 = 3.123
H ---> 4.20 / 1.008 = 4.167
O ---> 58.29 / 15.999 = 3.643

3) Look for lowest whole-number ratio:

C ---> 3.123 / 3.123 = 1
H ---> 4.167 / 3.123 = 1.334
O ---> 3.643 / 3.123 = 1.166

See that 1.334. That's one and one-third or 4/3. I'm going to multiply all three values by 3:

C ---> 1 x 3 = 3
H ---> 1.334 x 3 = 4
O ---> 1.166 x 3 = 3.5

See that 3.5? Let's now multiply through by 2.

C = 6
H = 8
O = 7

4) The empirical formula:

C6H8O7

When I found this question on an "answers" website, there was a wrong answer given:

C ---> 37.51/12 = 3.1258
H ---> 4.2/1 = 4.20
O ---> 58.29/16 = 3.6431
Mole proportion = CHO = Empirical formula.

Too much rounding off. Be very careful on rounding off on a problem like this citric acid one. Learn to recognize that something like 1.334 should be thought of as 4/3, leading to multiplying through by three. Do not round 1.334 off to 1 or round off something like 2.667 to three. And certainly, do not round off like the wrong-answer person did. No no no!

Problem #3: A compound is 19.3% Na, 26.9% S, and 53.8% O. Its formula mass is 238 g/mol. What is the molecular formula?

Solution:

1) We start by assuming 100 g of the compound is present. This turns the above percents into masses.

2) Calculate moles:

Na ---> 19.3 / 23.00 = 0.84
S ---> 26.9 / 32.1 = 0.84
O ---> 53.8 / 16.00 = 3.36

3) Look for lowest whole-number ratio:

3.36 / 0.84 = 4 (I only did the one for oxygen. You should be able to figure out the other two values!)

4) The empirical formula:

NaSO4

4) The molecular formula:

238 / 119 = 2

Na2S2O8

Problem #4: In which I present a problem and solution stripped down to their essentials. Hope you enjoy it!

C = 48.38%, H = 8.12%, O = 53.5%

Solution:

1) Moles of each:

4.028
8.06
3.34375

2) First attempt at a small whole-number ratio:

1.2
2.4
1

3) Multiply through by 10:

12
24
10

4) The empirical formula:

C6H12O5

Notice that I used a multiply by 10, then a divide by 2. You might ask: why not just multiply by 5? Well, you could, if you saw it. (Note: the ChemTeam did not see the multiply by five at first.) If you didn't, moving the decimal point to get whole numbers, then seeing the common factor gets you to the same place in a bit more round-about way.

That being said, if you saw that a multiply by five get you right to the empirical formula, then treat yourself to some ice cream!

Problem #5: Nitroglycerin has the following percentage composition:

carbon: 15.87%, hydrogen: 2.22%, nitrogen: 18.50%, oxygen: 63.41%

Determine its empirical formula.

Solution:

The assumption that 100 g of the compound is present turns the above percents into grams.

1) Calculate moles (I'll ignore units):

carbon ---> 15.87 / 12.01 = 1.321
hydrogen ---> 2.22/1.01 = 2.198
nitrogen ---> 18.50/14.01 = 1.320
oxygen ---> 63.41/16.0 = 3.963

2) Seek lowest whole-number ratio:

C ---> 1.321 / 1.32 = 1
H ---> 2.198 / 1.32 = 1.66
N ---> 1.32 / 1.32 = 1
O ---> 3.963 / 1.32 = 3

The key is the 1.66 which you do not round off to two. Think of it as 5/3.

3) Multiply everything by 3:

C ---> 1 x 3 = 3
H ---> 5/3 x 3 = 5
N ---> 1 x 3 = 3
O ---> 3 x 3 = 9

4) Empirical formula is:

C3H5N3O9

Problem #6: Insulin contains 3.4% sulphur. Calculate minimum molecular mass of insulin.

Solution:

1) Assume 100 g of insulin is present.

3.4% of that 100 grams is sulfur. Therefore, 3.4 g of sulfur is present.

2) Determine how many moles of sulfur are in 3.4 g of sulfur:

3.4 g / 32.065 g/mol = 0.106035 mol

3) Assume one mole of insulin contains one mole of sulfur:

 0.106035 100 g ––––––– = ––––––– 1 x

0.106035x = 100 g

x = 943 g

Problem #7: Two metallic oxides contain 27.6% and 30% oxygen in them respectively. If the formula of the first oxide is M3O4, then, what will be the formula of the second?

Solution:

Here you express everything on a per 100 g basis. The first oxide contains 27.6 g O or 27.6/16 = 1.725 moles of O and metal will be 100 - 27.6 = 72.4 g. Now the formula given is M3O4, so calculate 4 moles of O will react with how many g of metal, which will be 72.4*4/1.725 = 167.9 g of metal, which is equivalent to 3 moles of metal, so its atomic wt will be 167.9/3 = 55.97 or 56.

Do similar calculations for the second one, 30 g O = 30/16 = 1.875 moles reacting with 100-30 = 70 g metal. So the moles of metal will be 70/56 = 1.25 moles so the ratio of metal to oxygen is 1.25:1.875, divide by the smaller number which is 1.25, you get 1:1.5, you need to get to whole numbers, so it will be 2:3, therefore the formula will be M2O3

Problem #8: What formula yields 36.8% nitrogen in a nitrogen oxide?

Solution:

1) Write this:

 14N 0.368 = ––––––––––– 14N + 16O

Where N = the number of nitrogen atoms and O = the number of oxygen atoms

2) Cross multiply:

5.152N + 5.888O = 14N

3) Collect like terms:

5.888O = 8.848N

4) Divide through by smallest:

O = 1.5N

When N = 2, O = 3

5) The formula:
N2O3

6) Another way to think of it:

1.5N = O

3N = 2O

 3N 2O ––––––– = ––––––– (2) (3) (2) (3)

 N O ––––––– = ––––––– 2 3

N must equal 2 and O must equal 3 for the ratio and proportion to be equal.

Problem #9: A 150. g sample of a compound is found to be 44.1% C, 8.9% H and the remainder oxygen. What is the compound's empirical formula?

Solution:

1) Since percentages are given, we can assume 100 g (rather than 150 g) of compound is present:

C ---> 44.1 g
H ---> 8.9 g
O ---> 47.0 g (from 100 − 44.1 − 8.9)

Because the percentages are given, the fact that the sample is 150 g in mass is redundant.

2) Convert mass to moles:

C ---> 44.1 / 12.011 = 3.6716
H ---> 8.9/ 1.008 = 8.8294
O ---> 47.0 / 15.9994 = 2.9376

3) Use the smallest of answers above. Divide it into each answer:

C ---> 3.6716 / 2.9376 = 1.24986 mol = 1.25
H ---> 8.8294 / 2.9376 = 3.0056 = 3.00
O ---> 2.9376 / 2.9376 = 1.00

C ---> 1.25 = 54
H ---> 124
O ---> 44

5) Multiply through by 4:

C ---> 5
H ---> 12
O ---> 4

The empirical formula is C5H12O4

6) If your teacher were to insist on you using 150 g, then start this way:

C ---> (150) (0.441) =
H ---> (150) (0.089) =
O ---> (150) (0.470) =

and then convert the masses to moles and then do the calculations to get to the lowest set of whole-number subscripts.

Problem #10: Nitrogen forms more oxides than any other element. The percentage mass of nitrogen in one of the oxides is 36.85%.

(a) Determine the empirical formula of the compound
(b) Determine the molecular formula for this compound, given that its molecular weight is 152.0 g mol¯1

Solution: the typical way . . . .

1) Assume 100 g of compound is present.

2) Convert that %N and 100 g to mass N and mass O

N ---> 36.85 g
O ---> 100 − 36.85 = 63.15 g

3) Convert masses to moles

N ---> 36.85 g / 14.007 g/mol = 2.631 mol
O ---> 63.15 g / 15.9994 g/mol = 3.947 mol

4) Simplify mole ratio to get empirical formula.

N ---> 2.631 mol / 2.631 mol = 1
O ---> 3.947 mol / 2.631 = 1.5

Multiply by 2 to N = 2 and O = 3

N2O3 is the empirical formula

5) Compare molecular mass to empirical unit mass to get number of empirical units per molecule and thus molecular formula.

N2O3 weighs 76.0

152.0 / 76.0 = 2

N2O3 times 2 = N4O6

Solution: a different way . . . .

This method depends on knowing the molecular mass. if that value is not provided, we have to use the 'assume 100 g of the compound is present' method.

1) Determine the mass of N and O present in one mole of the nitrogen oxide:

N ---> (0.3685) (152.0 g) = 56.012 g
O ---> 152.0 − 56.012 = 95.988 g

The oxygen value could also be arrived at via this:

(0.6315) (152.0 g)

2) Determine moles of each:

N ---> 56.012 g / 14.007 g/mol = 3.9988577
O ---> 95.988 g / 15.9988 g/mol = 5.99969998

I think it's safe to round those answers off to 4 and 6.

3) Write the formulas:

molecular ---> N4O6
empirical ---> N2O3

Bonus Problem: An organic compound contains 45.90% C; 2.75% H; 7.65% N; 17.50% S; and the remainder is oxygen. (a) What is the empirical formula of the compound? (b) Knowing that the compound is saccharin, what is the molecular formula?

Comment: Note that the oxygen percentage is obtained by subtraction. The percent oxygen in a compound is difficult to determine experimentally, so all the other elements are determined by experiment and then oxygen is determined by subtraction.

Solution to (a):

1) Assume 100 g of the substance is present. This allows our percentages to be represented using grams:

C = 45.90 g; H = 2.75 g; N = 7.65 g; S = 17.50 g; O = 26.20 g

2) Determine moles of each element:

 45.90 g moles carbon = –––––––––––– =  3.8215 mol 12.011 g mol¯1

 2.75 g moles hydrogen = –––––––––––– =  2.728 mol 1.008 g mol¯1

 7.65 g moles nitrogen = –––––––––––– =  0.546 mol 14.007 g mol¯1

 17.50 g moles sulfur = –––––––––––– =  0.546 mol 32.065 g mol¯1

 26.20 g moles oxygen = –––––––––––– =  1.6376 mol 15.9994 g mol¯1

3) Divide through by the smallest value:

 3.8215 mol atoms carbon = –––––––––– =  6.999 = 7 0.546 mol

 2.728 mol atoms hydrogen = –––––––––– =  4.996 = 5 0.546 mol

 0.546 mol atoms nitrogen = –––––––––– =  1 0.546 mol

 0.546 mol atoms sulfur = –––––––––– =  1 0.546 mol

 1.6375 mol atoms oxygen = –––––––––– =  2.999 = 3 0.546 mol

4) The empirical formula is:

C7H5NO3S

Solution to (b):

1) We know the compound is saccharin, so we look up its molecular weight to be 183.15 g/mol.

2) The molecular formula is ALWAYS a whole number multiple of the empirical formula. That means that the molecular weight is ALWAYS a whole-number multiple of the empirical forula weight. Therefore:

183.15 / 183.15 = 1

The molecular formula is the same as the empirical formula.

The molecular formula is C7H5NO3S