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Determine the empirical formula for a compound containing 85.6 g of carbon and 14.4 g of hydrogen. Please show work
take gram & convert to mol (using the mm of the element) ,
85.6g C [tex]\frac{1 mol }{12.01g}[/tex] = 7.12 14.4g H[tex]\frac{1 mol}{1.01g}[/tex] =14.2 then take the smallest number & divide 7.12/7.12 =1 14.2/7.12= 2. since the numbers are in whole numbers the answer is [tex]C_{1}H_{2}[/tex].
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