1.367 g of an organic compound containing C,H,and O was combusted in a stream of air to yield 3.002g Co2 and 1.640 g H2O.What is its empirical formula ?
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Answer:
Let the empirical formula of the organic compound be represented by xC+yH+zO. We know it was heated in oxygen so we can represent the oxygen (O₂) by 2wO, where w, x, y, z are the numbers of atoms in the reagents.
The products can be represented by m(C+2O) and n(2H+O).
We need the atomic weights of the elements:
C=12.0096, O=15.9994, H=1.008.
Therefore we have 44.0084m and 18.0154n corresponding to 3.002g of CO₂ and 1.640g of H₂O. Some of the oxygen will be from the stream of oxygen (32w) and the rest from the organic compound (12.0096x+1.008y+15.9994z) (we’ll call this M) which means the total amount of oxygen is 31.9988w+15.9994z.
However, we know that there will be as many H and C atoms in the product as there were in the reagents, that is, in the organic compound, since the stream of O₂ was dry (contained no water). We can work out the mass of these elements in the product. 12.0096/44.0084×3.002g≃0.8192g of C in CO₂ and 1.008×1.640/18.0154≃0.09176g of H in H₂O. The amount of C and H in total is 0.8192+0.09176≃0.91096 or about 0.911g.
So the amount of O in the organic compound is about 1.367-0.911=0.456g.
12.0096x/M=0.8192/1.367=0.5993, so x/M=0.5993/12.0096≃0.05 because the ratio of C (12.0096x atoms) to the molecular weight of the organic compound is the same as the ratio of the amount (in grams) of C in the compound.
Also, 1.008y/M=0.0911/1.367≃0.0666, so y/M≃0.0661, and
15.9994z/ M=0.456/1.367≃0.3336, so z/M≃0.0208 for H and O.
We only need the ratios C:H:O to establish the empirical formula, so we need x:y:z.
x:y=0.05÷0.0666=0.75=3/4 approx. C:H=3:4.
y:z=0.0666÷0.0208=3.20 approx=16/5 or perhaps 3.33=10/3.
So C:H:O=3:4:(4×16/5)=3:4:12.8 or 3:4:(4×10/3)=3:4:40/3.
So multiply through by 5: 15:20:64=C₁₅H₂₀O₆₄, or
multiply by 3: 9:12:40=C₉H₁₂O₄₀.