0.3194 g of C, 0.05361 g of H and 2.125 g Br
Convert grams to mols.
mols C = 0.3194/12.01 = approx 0.0265
mols H = 0.05361/1.008 = about 0.0536
mols Br = 2.125/79.9 = about 0.0265
Now you want to find the ratio of each element to the others with the smallest number being 1 and all small whole numbers. The easy way to do that is to divide the smallest number by itself, then divide the others number by the same small number.
C = 0.0265/0.0265 = 1.00
H = 0.05361/0.0265 = 2.02
Br = 0.0265/0.0265 = 1.00
Rounding the 2.02 to 2.00 the empirical formula is C1H2O1 or CH2O.
If the molecular compound is known to contain 12 H atoms, just multiply everything by 6 to get C6H12O6 for the molecular formula.
This problem involves a molecular compound so both empirical and molecular formulas are required. A compound composed of Carbon, Hydrogen, and Bromine is analyzed. It is found to contain 0.3194 g of C, 0.05361 g of H and 2.125 g Br. What is the empirical formula of the compound? If the compound is known to have 12 H atoms in it, what is the molecular formula of the compound?
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