An organic compound on analysis gave the following compositions carbon =40% H=6 6% o=53% calculate its molecular formula if its molecular mass 90
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In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers.
Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.
So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.
Clearly n=2, and the molecular formula is 2×(CH2O
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers.
Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.
So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.
Clearly n=2, and the molecular formula is 2×(CH2O
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