SOME OTHER INTERPRETATIONS OF CHEMICAL
FORMULASOnce we master the mole concept and the meaning of chemical formulas, we can use them
in many other ways. The examples in this section illustrate a few additional kinds of in-
formation we can get from a chemical formula and the mole concept.
EXAMPLE 2-18 Composition of Compounds
What mass of chromium is contained in 35.8 g of (NH 4 ) 2 Cr 2 O 7?
Plan
Let us first solve the problem in several steps.
Step 1: The formula tells us that each mole of (NH 4 ) 2 Cr 2 O 7 contains two moles of Cr atoms,
so we first find the number of moles of (NH 4 ) 2 Cr 2 O 7 , using the unit factor
Step 2: Then we convert the number of moles of (NH 4 ) 2 Cr 2 O 7 into the number of moles of
Cr atoms it contains, using the unit factor
Step 3: We then use the atomic weight of Cr to convert the number of moles of chromium
atoms to mass of chromium.
Mass (NH 4 ) 2 Cr 2 O 7 88n mol (NH 4 ) 2 Cr 2 O 7 88n mol Cr 88n Mass CrSolution
Step 1: ? mol (NH 4 ) 2 Cr 2 O 7 35.8 g (NH 4 ) 2 Cr 2 O 7
0.142 mol (NH 4 ) 2 Cr 2 O 7Step 2: ? mol Cr atoms0.142 mol (NH 4 ) 2 Cr 2 O 7
0.284 mol Cr atomsStep 3: ? g Cr0.284 mol Cr atoms14.8 g Cr
If you understand the reasoning in these conversions, you should be able to solve this prob-
lem in a single setup:
52.0 g Cr
1 mol Cr atoms2 mol Cr atoms
1 mol (NH 4 ) 2 Cr 2 O 71 mol (NH 4 ) 2 Cr 2 O 7
252.0 g (NH 4 ) 2 Cr 2 O 72 mol Cr atoms
1 mol (NH 4 ) 2 Cr 2 O 71 mol (NH 4 ) 2 Cr 2 O 7
252.0 g (NH 4 ) 2 Cr 2 O 72-11
2-11 Some Other Interpretations of Chemical Formulas 77? g Cr35.8 g (NH 4 ) 2 Cr 2 O 7 14.8 g Cr
52.0 g Cr
1 mol Cr2 mol Cr atoms
1 mol (NH 4 ) 2 Cr 2 O 71 mol (NH 4 ) 2 Cr 2 O 7
252.0 g (NH 4 ) 2 Cr 2 O 7You should now work Exercise 72.