Figure 1-21. Left to right: John Dalton (1766-1844) (source), Thomas Thomson (1773-1852) (source), Jöns Jakob Berzelius (1779-1848) (source), and Stanislao Cannizzaro (1826-1910) (source).
Although herein we skip details of how the atomic masses were determined in the original studies, one example below can give you an idea. Knowing the composition of a compound comprising two elements and the atomic mass of one of the two, one can determine the atomic mass of the other element. For example, methane, the main component of natural gas, has the formula CH4
. It was determined experimentally that 100 g of methane consists of 25 g of hydrogen and 75 g of carbon. Given the atomic weight of hydrogen (1 a.m.u., originally assigned by John Dalton), we can calculate the atomic weight of carbon. The mass carbon to hydrogen ratio (75 : 25 = 3) shows that one
C atom is 3 times heavier than four
H atoms. Why four? Because in a molecule of methane, there are four hydrogen atoms per each carbon atom, as indicated by the formula, CH4
. Since the mass of four hydrogen atoms equals 4 x 1 a.m.u. = 4 a.m.u., the atomic mass of carbon is 3 x 4 a.m.u. = 12 a.m.u. 1.7.3. Exercises.
1. Calculate mass percentages of each element in the following compounds: (a) CaO, (b) Al2
, (c) MgBr2
, (d) CuSO4
, (e) CuSO4
2. You want to invest in gold. There is a vendor selling gold trichloride (AuCl3
) and gold tribromide (AuBr3
). The price per gram is $100.00 for AuBr3
and $140.00 for AuCl3
. Should you buy AuCl3
to extract more value from your purchase? Neglect the value of the other two elements, Cl and Br. Answer