Question 6.1: Calculate the formula mass of each of the following substanc......

Formula masses are obtained simply by adding the atomic masses of the constituent elements, counting each atomic mass as many times as the symbol for the element occurs in the chemical formula.
a. A formula unit of SnF_{2} contains three atoms: one atom of Sn and two atoms of F. The formula mass, the collective mass of these three atoms, is calculated as follows:
\quad\quad\quad\quad 1  \cancel{atom  Sn}\times (\frac{118.71  amu}{1  \cancel{atom  Sn}})=118.71  amu
\quad\quad\quad\quad\quad 2  \cancel{atom  F}\times (\frac{19.00  amu}{1  \cancel{atom  F}})=\underline{38.00  amu}
\quad\quad\quad\quad\quad\quad\quad \ \quad Formula mass = 156.71 amu
We derive the conversion factors in the calculation from the atomic masses listed on the inside front cover of the text. Our rules for the use of conversion factors are the same as those discussed in Section 2.7.
\quad\quadConversion factors are usually not explicitly shown in a formula mass calculation, as they are in the preceding calculation; the calculation is simplified as follows:
\quad\quad\quad\quadSn:     1 × 118.71 amu = 118.71 amu
\quad\quad\quad\quadF:       2 × 19.00 amu = \underline{38.00  amu}
\quad\quad\quad\quad         Formula mass = 156.71 amu
b. The chemical formula for this compound contains parentheses. Improper interpretation of parentheses (see Section 4.11) is a common error made by students doing formula mass calculations. In the formula Al(OH)_{3}, the subscript 3 outside the parentheses affects both of the symbols inside the parentheses. Thus we have
\quad\quad\quad\quadAl:     1 × 26.98 amu = 26.98 amu
\quad\quad\quad\quadO:      3 × 16.00 amu = 48.00 amu
\quad\quad\quad\quadH:         3 × 1.01 amu = \underline{3.03  amu}
\quad\quad\quad\quad         Formula mass= 78.01 amu
In this text, we will always use atomic masses rounded to the hundredths place, as we have done in this example. This rule allows us to use, without rounding, the atomic masses given inside the front cover of the text. A benefit of this approach is that we always use the same atomic mass for a given element and thus become familiar with the atomic masses of the common elements.