Write two molecular formulas that reduce to each to each of the given empirical formulas. Do not include the given empirical formulas.
1. Al2(SO4)3
. Cr(CN)3
Cr2(CN)6 ?
Al4(SO4)6 ?
To find the molecular formulas that reduce to the given empirical formulas, we need to determine the number of atoms of each element in the empirical formula and then find a multiple of these numbers that gives us a whole number ratio.
1. Al2(SO4)3:
The empirical formula Al2(SO4)3 tells us that there are 2 Al (aluminum) atoms, 3 S (sulfur) atoms, and 12 O (oxygen) atoms.
To find the molecular formula, we need to find a multiple of these numbers that gives us a whole number ratio. Let's assume the molecular formula is AlxSxOx.
The ratio of Al to S to O in the empirical formula is 2:3:12. To find the least common multiple (LCM) of these numbers, we have:
Al: 2, 4, 6, 8, 10, 12, 14, 16, ...
S: 3, 6, 9, 12, 15, 18, ...
O: 12, 24, ...
The LCM of 2, 3, and 12 is 12. So, the molecular formula that reduces to Al2(SO4)3 is Al12(S12O48).
2. Cr(CN)3:
The empirical formula Cr(CN)3 tells us that there is 1 Cr (chromium) atom, 3 C (carbon) atoms, and 3 N (nitrogen) atoms.
Let's assume the molecular formula is CrxCyNz.
The ratio of Cr to C to N in the empirical formula is 1:3:3. To find the LCM of these numbers, we have:
Cr: 1, 2, 3, 4, 5, 6, 7, ...
C: 3, 6, ...
N: 3, 6, ...
The LCM of 1 and 3 is 3. So, the molecular formula that reduces to Cr(CN)3 is Cr3(C9N9).