Empirical Formulas
Suppose you analyze an unknown compound that is a white powder and find that it is composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. You can use those percentages to determine the mole ratios among sodium, sulfur, and oxy-gen and write a formula for the compound.
To begin, the mass percentages of each element can be interpreted as “grams of element per 100 grams of compound.” To make things simpler, you can assume you have a 100 g sample of the unknown compound. The unknown compound contains 36.5% sodium by mass. Therefore 100.0 g of the compound would con-tain 36.5 g of sodium. You already know how to convert mass of a substance into number of moles, so you can calculate the number of moles of sodium in 36.5 g. After you find the number of moles of each element, you can look for a simple ratio among the elements and use this ratio of elements to write a formula for the compound.
The chemical formula obtained from the mass percentages is in the simplest form for that compound. The mole ratios for each element, which you deter-mined from the analytical data given, are reduced to the smallest whole numbers. This simplest formula is also called the empirical formula. The actual formula for the compound could be a multiple of the empirical formula. For instance, sup-pose you analyze a compound and find that it is comsup-posed of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. If you determine the formula for this compound based only on the analytical data, you will determine the formula to be CH2O. There are, however, other possibilities for the formula. It could be C2H4O2and still have the same percentage composition. In fact, it could be any multiple of CH2O.
It is possible to convert from the empirical formula to the actual chemical for-mula for the compound as long as the molar mass of the compound is known. Look again at the CH2O example. If the true compound were CH2O, it would have a molar mass of 30.03 g/mol. If you do more tests on the unknown compound and find that its molar mass is 60.06, you know that CH2O cannot be its true identity. The molar mass 60.06 is twice the molar mass of CH2O. Therefore, you know that the true chemical formula must be twice the empirical formula, (CH2O) 2, or C2H4O2. Any correct molecular formula can be determined from an empirical for-mula and a molar mass in this same way.
General Plan for Determining Empirical Formulas and Molecular Formulas
Convert using the molar mass of each element.
The calculated ratio is the simplest formula. Percentage of element expressed as grams of element per 100 g unknown Empirical formula of the compound Calculated whole-number ratio among the
elements Use the amount of the least-abundant element to calculate the simplest whole-number ratio among the elements.
Convert using the experimental molar mass of the unknown and the molar mass of the simplest formula. 1 4 Molecular formula of the compound 5 Amount of each element per 100 g of unknown 2 3
Sample Problem 1
Determine the empirical formula for an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur by mass.
Solution
ANALYZEWhat is given in the problem? the percentage composition of the compound
What are you asked to find? the empirical formula for the compound
PLAN
What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown?
State the percentage of the element in grams and multiply by the inverse of the molar mass of the element.
What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)?
Divide the amount of each element by the amount of the least-abundant ele-ment. If necessary, multiply the ratio by a small integer that will produce a whole-number ratio.
multiply by the inverse of the molar mass of Na
Mass of Na per 100.0 g unknown
1
Amount Na in mol per 100.0 g unknown
2
1 molar mass Na percent of Na stated as grams
Na per 100 g unknown 1 mol Na 22.99 g Na 36.5 g Na 100.0 g unknown mol Na 100.0 g unknown Items Data
The percentage composition of the unknown subtance 36.5% sodium 38.1% oxygen 25.4% sulfur The molar mass of each element* 22.99 g Na/mol Na
16.00 g O/mol O 32.07 g S/mol S Amount of each element per 100.0 g of the unknown ? mol
Simplest mole ratio of elements in the unknown ?
Repeat this step for the remaining elements.
COMPUTE
Divide the amount of each element by the amount of the least-abundant ele-ment, which in this example is S. This can be accomplished by multiplying the amount of each element by the inverse of the amount of the least abundant ele-ment.
From the calculations, the simplest mole ratio is 2 mol Na : 3 mol O : 1 mol S. The simplest formula is therefore Na2O3S. Seeing the ratio 3 mol O : 1 mol S, you can use your knowledge of chemistry to suggest that this possibly represents a sulfite group, –SO3and propose the formula Na2SO3.
EVALUATE
Are the units correct?
Yes; units canceled throughout the calculation, so it is reasonable to assume that the resulting ratio is accurate.
Is the number of significant figures correct?
Yes; ratios were calculated to three significant figures because percentages were given to three significant figures.
100.0 g unknown 0.792 mol S 0.792 mol S 100.0 g unknown 1.00 mol S 1 mol S 100.0 g unknown 0.792 mol S 2.38 mol O 100.0 g unknown 3.01 mol O 1 mol S 100.0 g unknown 0.792 mol S 1.59 mol Na 100.0 g unknown 2.01 mol Na 1 mol S 1 mol S 32.07 g S 25.4 g S 100.0 g unknown 0.792 mol S 100.0 g unknown 1 mol O 16.00 g O 38.1 g O 100.0 g unknown 2.38 mol O 100.0 g unknown 1 mol Na 22.99 g Na 36.5 g Na 100.0 g unknown 1.59 mol Na 100.0 g unknown divide by the amount of the least-abundant element
Amount of Na in mol per 100.0 g unknown
2
4 Whole-number ratio
among the elements
3
Is the answer reasonable?
Yes; the formula, Na2SO3 is plausible, given the mole ratios and considering that
the sulfite ion has a 2 charge and the sodium ion has a 1 charge.
Practice
1.Determine the empirical formula for compounds that have the following analyses:
a.28.4% copper, 71.6% bromine ans: CuBr2
b.39.0% potassium, 12.0% carbon, 1.01% hydrogen, and 47.9% oxygen
ans: KHCO3
c.77.3% silver, 7.4% phosphorus, 15.3% oxygen ans: Ag3PO4
Sample Problem 2
Determine the empirical formula for an unknown compound composed of 38.4% oxygen, 23.7% carbon, and 1.66% hydrogen.
Solution
ANALYZEWhat is given in the problem? the percentage composition of the compound
What are you asked to find? the empirical formula for the compound
PLAN
What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown?
State the percentage of the element in grams and multiply by the inverse of the molar mass of the element.
What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)?
Divide the amount of each element by the amount of the least-abundant ele-ment. If necessary, multiply the ratio by a small integer to produce a whole-number ratio.
COMPUTE
Proceed to find the amount in moles per 100.0 g of unknown for the elements carbon, oxygen, and hydrogen, as in Sample Problem 1.
When determining the formula of a compound having more than two elements, it is usually advisable to put the data and results in a table.
1 mol K 39.10 g K 38.4 g K 100.0 g unknown 0.982 mol K 100.0 g unknown
divide by the amount of the least-abundant element, and multiply by an integer that will produce a whole-number ratio multiply by the inverse of
the molar mass of K
Mass of K in g per 100.0 g unknown
1
Empirical formula
4 Whole-number ratio
among the elements
3
Amount of K in mol per 100.0 g unknown
Again, as in Sample Problem 1, divide each result by the amount in moles of the least-abundant element, which in this example is K.
You should get the following results:
In contrast to Sample Problem 1, this calculation does not give a simple whole-number ratio among the elements. To solve this problem, multiply by a small integer that will result in a whole-number ratio. You can pick an integer that you think might work, or you can convert the number of moles to an equivalent fractional number. At this point, you should keep in mind that analytical data is never perfect, so change the number of moles to the fraction that is closest to the decimal number. Then, choose the appropriate integer factor to use. In this case, the fractions are in thirds so a factor of 3 will change the fractions into whole numbers.
Thus, the simplest formula for the compound is K3C6H5O7, which happens to be the formula for potassium citrate.
Amount in mol of element Amount in mol of element Element per 100.0 g of unknown per mol of potassium
Potassium 0.982 mol K 1.00 mol K
Carbon 1.97 mol C 2.01 mol C
Oxygen 2.27 mol O 2.31 mol O
Hydrogen 1.64 mol H 1.67 mol H
Amount in mol
of element per Fraction nearest the Integer Whole-number mole of potassium decimal value factor mole ratio
1.00 mol K 1 mol K 3 3 mol K
2.01 mol C 2 mol C 3 6 mol C
2.31 mol O 2 1/3 mol O 3 7 mol O
1.67 mol H 1 2/3 mol H 3 5 mol H
Mass per 100.0 g Amount in mol per Element of unknown Molar mass 100.0 g of unknown
Potassium 38.4 g K 39.10 g/mol 0.982 mol K
Carbon 23.7 g C 12.01 g/mol 1.97 mol C
Oxygen 36.3 g O 16.00 g/mol 2.27 mol O
EVALUATE
Is the answer reasonable?
Yes; the formula, K3C6H5O7is plausible, considering that the potassium ion has
a 1 charge and the citrate polyatomic ion has a 3 charge.
Practice
1.Determine the simplest formula for compounds that have the following analy-ses. The data may not be exact.
a.36.2% aluminum and 63.8% sulfur ans: Al2S3
b.93.5% niobium and 6.50% oxygen ans: Nb5O2
c.57.6% strontium, 13.8% phosphorus, and 28.6% oxygen ans: Sr3P2O8or
Sr3(PO4)2
Sample Problem 3
A compound is analyzed and found to have the empirical formula CH2O.
The molar mass of the compound is found to be 153 g/mol. What is the compound’s molecular formula?
Solution
ANALYZEWhat is given in the problem? the empirical formula, and the experimental molar mass
What are you asked to find? the molecular formula of the compound
PLAN
What steps are needed to determine the molecular formula of the unknown com-pound?
Multiply the experimental molar mass by the inverse of the molar mass of the empirical formula. The subscripts of the empirical formula are multiplied by the whole-number factor obtained.
COMPUTE
Allowing for a little experimental error, the molecular formula must be five times the empirical formula.
5.09 mol CH2O 1 mol unknown 1 mol CH2O 30.03 g 153 g 1 mol unknown
multiply the experimental molar mass by the inverse of the molar mass of the empirical formula, and multiply each subscript in the empirical formula by
the resulting factor
Empirical formula of unknown 4 Molecular formula of unknown 5 1 molar mass of empirical formula
factor that shows the number of times the empirical formula must be multiplied to get the
molecular formula given 1 mol CH2O 30.03 g 153 g 1 mol unknown mol CH2O 1 mol unknown Items Data
Empirical formula of unknown CH2O
Experimental molar mass of unknown 153 g/mol Molar mass of empirical formula 30.03 g/mol Molecular formula of the compound ?
EVALUATE
Is the answer reasonable?
Yes; the calculated molar mass of C5H10O5is 150.15, which is close to the
exper-imental molar mass of the unknown. Reference books show that there are sev-eral different compounds with the formula C5H10O5.
Practice
1.Determine the molecular formula of each of the following unknown sub-stances:
a.empirical formula CH2, experimental molar mass 28 g/mol ans: C2H4
b.empirical formula B2H5, experimental molar mass 54 g/mol ans: B4H10
c.empirical formula C2HCl, experimental molar mass 179 g/mol ans: C6H3Cl3
d.empirical formula C6H8O, experimental molar mass 290 g/molans: C18H24O3
Additional Problems
1.Determine the empirical formula for compounds that have the following analyses:
a.66.0% barium and 34.0% chlorine
b.80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen
c.12.67% aluminum, 19.73% nitrogen, and 67.60% oxygen
d.35.64% zinc, 26.18% carbon, 34.88% oxygen, and 3.30% hydrogen
e.2.8% hydrogen, 9.8% nitrogen, 20.5% nickel, 44.5% oxygen, and 22.4% sulfur
f. 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine
2.Sometimes, instead of percentage composition, you will have the composition of a sample by mass. Use the same method shown in Sample Problem 1, but use the actual mass of the sample instead of assuming a 100 g sample. Determine the empirical formula for compounds that have the following analyses:
a.a 0.858 g sample of an unknown substance is composed of 0.537 g of cop-per and 0.321 g of fluorine
b.a 13.07 g sample of an unknown substance is composed of 9.48 g of barium, 1.66 g of carbon, and 1.93 g of nitrogen
c.a 0.025 g sample of an unknown substance is composed of 0.0091 g man-ganese, 0.0106 g oxygen, and 0.0053 g sulfur
3.Determine the empirical formula for compounds that have the following analyses:
a.a 0.0082 g sample contains 0.0015 g of nickel and 0.0067 g of iodine
b.a 0.470 g sample contains 0.144 g of manganese, 0.074 g of nitrogen, and 0.252 g of oxygen
c.a 3.880 g sample contains 0.691 g of magnesium, 1.824 g of sulfur, and 1.365 g of oxygen
d.a 46.25 g sample contains 14.77 g of potassium, 9.06 g of oxygen, and 22.42 g of tin
4.Determine the empirical formula for compounds that have the following analyses: a.60.9% As and 39.1% S b.76.89% Re and 23.12% O c.5.04% H, 35.00% N, and 59.96% O d.24.3% Fe, 33.9% Cr, and 41.8% O e.54.03% C, 37.81% N, and 8.16% H f. 55.81% C, 3.90% H, 29.43% F, and 10.85% N
5.Determine the molecular formulas for compounds having the following empir-ical formulas and molar masses:
a.C2H4S; experimental molar mass 179 b.C2H4O; experimental molar mass 176 c.C2H3O2; experimental molar mass 119 d.C2H2O, experimental molar mass 254
6.Use the experimental molar mass to determine the molecular formula for compounds having the following analyses:
a.41.39% carbon, 3.47% hydrogen, and 55.14% oxygen; experimental molar mass 116.07
b.54.53% carbon, 9.15% hydrogen, and 36.32% oxygen; experimental molar mass 88
c.64.27% carbon, 7.19% hydrogen, and 28.54% oxygen; experimental molar mass 168.19
7.A 0.400 g sample of a white powder contains 0.141 g of potassium, 0.115 g of sulfur, and 0.144 g of oxygen. What is the empirical formula for the com-pound?
8. A 10.64 g sample of a lead compound is analyzed and found to be made up of
9.65 g of lead and 0.99 g of oxygen. Determine the empirical formula for this compound.
9.A 2.65 g sample of a salmon-colored powder contains 0.70 g of chromium, 0.65 g of sulfur, and 1.30 g of oxygen. The molar mass is 392.2. What is the formula of the compound?
10.Ninhydrin is a compound that reacts with amino acids and proteins to pro-duce a dark-colored complex. It is used by forensic chemists and detectives to see fingerprints that might otherwise be invisible. Ninhydrin’s composition is 60.68% carbon, 3.40% hydrogen, and 35.92% oxygen. What is the empirical formula for ninhydrin?
11.Histamine is a substance that is released by cells in response to injury, infec-tion, stings, and materials that cause allergic responses, such as pollen. Histamine causes dilation of blood vessels and swelling due to accumulation of fluid in the tissues. People sometimes take antihistamine drugs to counter-act the effects of histamine. A sample of histamine having a mass of 385 mg is composed of 208 mg of carbon, 31 mg of hydrogen, and 146 mg of nitrogen. The molar mass of histamine is 111 g/mol. What is the molecular formula for histamine?
12.You analyze two substances in the laboratory and discover that each has the empirical formula CH2O. You can easily see that they are different substances because one is a liquid with a sharp, biting odor and the other is an odorless, crystalline solid. How can you account for the fact that both have the same empirical formula?
