CHEMICAL FORMULAS
Topics on this page: Empirical Formulas, Molecular Formulas, % Composition, Problems and Answers\

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Chemical formulas are either empirical or molecular. The empirical formula represents the simplest or smallest whole number ratio of atoms while the molecular (or true) formula gives the actual number of atoms of each element in one molecule.

CALCULATION OF EMPIRICAL FORMULAS

The empirical formula of a compound may be determined from a knowledge of the composition by weight and the atomic weights of the elements represented in the compound.

Step 1: From the percentage by weight, calculate the number of moles of each element present in 100 g of the substance.

Step 2: Obtain the simplest whole number ration of moles of each element. This is done by dividing the number of moles of each element in Step 1 by the smallest number of moles calculated.

Example 1: A certain hydrocarbon contains 25% hydrogen and 75% carbon by weight. Calculate its empirical formula.

Solution: From a table of atomic weights: H = 1.0 and C = 12.0. Set up a tabular form:

Empirical formula is CH₄

The student should note that in Step 2, dividing through by the smallest quotient does not always give a simple whole number ratio.  Whole number ratios must exist because after all we have calculated relative numbers of moles and moles refer to numbers of atoms. Any ratios which are not calculated to be whole numbers must be attributed to experimental error arising from the % determinations.  If this is the case, it must not be rounded off, but instead multiplied through by the smallest factor necessary to give whole numbers. The following example illustrates this point.

Example 2: A compound contains 6.841 g of chromium combined with 3.158 g of oxygen. Calculate its empirical formula.

Solution: Atomic weights, Cr = 52.0 and 0 = 16.0

Empirical formula is Cr₂O₃

PERCENT COMPOSITION FROM EMPIRICAL FORMULA

The percent composition by weight of a compound can be determined from its chemical formula.

Example 3: Calculate the percentage composition of K₂CO₃ .

Solution: K = 39.1, C = 12.0, and O = 16.0

Step 1: Calculate relative weight of each element and formula weight.

Step 2: Divide the relative weight of each element by the formula weight. Multiply by 100 to convert to percent.

Total % of all elements must add to 100%

CALCULATION OF MOLECULAR FORMULAS

If, in addition to composition data, the molecular weight is known or data by which the molecular weight can be calculated is available, then the true or molecular formula of the substance may be determined.

Example 4: A certain gas contains 85.7% carbon and 14.3% hydrogen. Its molecular weight is 42. Determine its molecular formula.

Empirical formula is CH₂. Empirical formula weight = 12 + 2 = 14.

; therefore the molecular formula is (CH2)·3 or C3H6

PROBLEMS

1. A certain phosphate of magnesium has the following composition: Mg = 21.85%, P = 27.85%, and 0 50.31%. Determine its simplest formula.

2. Calculate the empirical formula of the compound known to contain 0.7888 g of gold combined with 0.4260 g of chlorine.

3. How many ounces of silver can be obtained from a pound of AgNO₃ ?

4. How many grams of oxygen are there in 100.0 g of 903% pure KCO₃ ?

5. Determine the true formula of the compound having a molecular weight of approximately 92 and containing 0.36 g of carbon, 0.090 g of hydrogen, and 0.24 g of oxygen.

6. The empirical formula of a certain compound is AB₂. If it is known to contain 60% A, having an atomic weight of 30 what must be the atomic weight of B?

7. A sample of hydrocarbon weighing 0.5000 g produced 1.650 g of CO₂ when burned in pure oxygen. What is its empirical formula?

8. Calculate the empirical formula of a mineral if chemical analysis showed the following composition: 20.0% CaO, 15.7% MgO, and 64.3% SiO₂

9. Hydrated calcium phosphite contains 81.63% anhydrous CaHPO_3. Derive its simplest formula.

10. Calculate the simplest formulas of the compounds formed as indicated below.

(a) 1.1817 g Cu combined with 1.3183 g Cl

(b) 1.2432 g Pb combined with 0.1280 g O

(c) 46.55% Fe and 53.45%. S

ANSWERS TO PROBLEMS

1. Mg₂P₂O₇

2. AuCl₃

3. 10.16 oz.

4. 35.2

5. C₄H₁₂O₂

6. 10

7. C₃H₄

8. CaMgSi₃O₈

9. CaHPO₃ ·3H₂O

10. (a) CuCl₂ (b) Pb₃O₄ (c) FeS₂