@CBMiii3,
Are you talking about by weight or by mole--if by mole you have to use molecular weight--take the first one
Quote:What is the percentage composition of an ore which contains 1.85 g of aluminum
and 1.65 g of Oxygen?
By weight---total , T=1.85g+1.65g=3.50g
% Al (by weight)=1.85g/3,50g*100?=~52.9% (by weight)
% O (by weight)=1.65g/3.5g*100%=~47.1% (by weight)
Now if you want molar ratios
mw Al=~27g/mol
mw O=~16g/mol
Tale 100g of ore
53gAl/27gAl/mol=1.96molAl=~2molAl
47g)/16g)/mol=2.93molO=~3molO
The molar ratios are 2 to 3, s the empirical formulae for this ore is Al2O3.
Te other three problems can be solved the same way.
The only additional thing you'll need is this--you should have one in your science book.
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