Hi there in this question, we are given molecular formulas of c 2, h, 6 c, 3, h, 6 c, 2, h, 2 and c 6 h, 6, and the first thing we want to determine is the empirical formula for each 1. Well, the empirical formula, as it’s mentioned in the direction says it’s simply: the chemical formula reduced to the lowest whole number ratio. So when we have c 2 h, 62 and 6 are both divisible by 2, so we’re reducing. So if we divide each of these by 2, we get c h, 3, technically, it’s c 1, but we don’t write the 1. So it’s ch 3 for c 3. H 6. We see that both of those numbers are divisible by 3. So if we divide those both by 3, we get c h, 2 for c 2, h, 2, they’re both divisible by 2. So we simply get an empirical formula of c h, o 1 to 1 ratio, and the same is true for c 6 h. 6. It’S a 1 to 1 ratio. Well, when we figure out the percent composition, we can use the empirical formula because the molecular formula is just a multiple of that, so the percentage is not going to change between the molecular formula and the empirical formula. So i am just going to use the empirical formula to determine our percent carbon and our percent hydrogen and to determine percent composition. We need to take the mass of the compound we’re interested in so in this case of carbon, it’s the carbon there’s just 1 carbon. So it’s going to…

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