Calculate empirical formula when given percent composition data

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Calculate 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 the formula of a hydrate


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:"

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 minus 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 divided by 6.022 x 1023 atoms/mol = 0.071903 mol

0.071903 mol times 16.00 g/mol = 1.15045 g

1.15045 g / 3.25 g = 0.3540 = 35.40%

2) Percent chlorine:

100 minus (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.


Example #11: 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


Example #12: 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!


Example #13: 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


Example #14: 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!


Example #15: 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

Example #16: 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 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


Example #17: 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:

I will reproduce the answer given on an "answers" website:

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


Example #18: 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.


Example #19: 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

4) Think about the answers from step 3 as improper fractions:

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.


Example #20: 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

Example #21: 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

Example #22: 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

Example #23: 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
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
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


Bonus Example #1: 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.


Bonus Example #2: 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.


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