Originally posted by: virtueixi
A compound contains is consists of the following
52.14 % of C
13.13 % of H
34.73 % of O
If a 0.08 mole sample of this compound weighs 7.37 g
what is the molecular formula of this compound?
C=12.01 H=1.008 O= 16 (molar masses)
thx .
First I'd like to say the question isn't written very well. Is 52.14% of the total mass Carbon? Or is that 52.14% of the total atoms? I'll assume it refers to mass.
Lets start with the easy part: 7.37 g / 0.08 mol = 92.125 g / mol. So we know the molar weight of the compound.
Now lets look at those percentages. Suppose we reach on in and grab 92.125 g of atoms. On average if we could randomly pick atoms we'd have picked 48.03 g of C atoms (92.125 g *52.14%), 12.10 g of H atoms (92.125 g *13.13%), and 32.00 g of O atoms (92.125 g *34.73%). How many moles is that of each atom? 48.03 g C / (12.01 g/mol) = 4.00 mol of C. 12.10 g H / (1.008 g/mol) = 12.00 mol of H. 32.00 g O / (16 g/mol) = 2.00 mol O.
So your molecule has 4 C atoms, 12 H atoms, and 2 O atoms.
Check: 4*12.01 + 12*1.008 + 2*16 = 92.14 (close enough to the true molecular weight due to rounding errors I made).
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