find the moles of chromium ... (sample mass) / (molar mass)
find the mass of oxygen ... (chromium oxide mass) - (chromium metal mass)
find the moles of oxygen ... (mass of oxygen) / (molar mass of oxygen)
find the LCM of the moles
... empirical formula is the ratio of the moles
An analysis shows that 5.85g of chromium metal combines with oxygen to form 8.62g of chromium oxide. What is the empirical formula of the compound? Please help!!
2 answers
I do this a slightly different way than R_Scott as follows:
% Cr = (5.85/8.62)*100 = about 67.86
% O = 100% - 67.86 = about 32.14
g Cr = 67.86
g O = 32.14
mols Cr = 67.86/52 = about 1.3
mols O = 32.14/16 = about 2.0
Divide both numbers by the smaller; i.e., by 1.3 like this
1.3/1.3 = 1
2.0/1.3 = 1.54 and that 4 is not significant.
You want whole numbers so multiply by an integer to obtain whole numbers. Multiplying by 2 gives Cr = 2 and O = 3.08 and this rounds to 2:3 so the empirical formula is Cr2O3.
% Cr = (5.85/8.62)*100 = about 67.86
% O = 100% - 67.86 = about 32.14
g Cr = 67.86
g O = 32.14
mols Cr = 67.86/52 = about 1.3
mols O = 32.14/16 = about 2.0
Divide both numbers by the smaller; i.e., by 1.3 like this
1.3/1.3 = 1
2.0/1.3 = 1.54 and that 4 is not significant.
You want whole numbers so multiply by an integer to obtain whole numbers. Multiplying by 2 gives Cr = 2 and O = 3.08 and this rounds to 2:3 so the empirical formula is Cr2O3.
1 answer