Empirical to molecular
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Re: Empirical to molecular
You would have to be given the molar mass (which is usually said to be done through mass spectrometry).
In this example we did in class: Vitamin C sample of 8.00g is analyzed - the results are C 3.27g, H 0.366g, O 4.36g
Let's jump down the steps and assume that you have calculated that the empirical formula for Vitamin C would be C3H4O3.
Then, you would have to be given the molar mass. Let's say we are told it is 176.14 g.mol-1. Since the molar mass of C3H4O3 is 88.06 g.mol-1, you would then divide 176.14 g.mol-1 by 88.06g.mol-1 to get 2.00.
So, you then multiply 2 by each "coefficient" of the elements, so C3H4O3 becomes C6H8O6.
C3H4O3 is the empirical formula and C6H8O6 is the molecular formula for Vitamin C.
Hope this helped!
In this example we did in class: Vitamin C sample of 8.00g is analyzed - the results are C 3.27g, H 0.366g, O 4.36g
Let's jump down the steps and assume that you have calculated that the empirical formula for Vitamin C would be C3H4O3.
Then, you would have to be given the molar mass. Let's say we are told it is 176.14 g.mol-1. Since the molar mass of C3H4O3 is 88.06 g.mol-1, you would then divide 176.14 g.mol-1 by 88.06g.mol-1 to get 2.00.
So, you then multiply 2 by each "coefficient" of the elements, so C3H4O3 becomes C6H8O6.
C3H4O3 is the empirical formula and C6H8O6 is the molecular formula for Vitamin C.
Hope this helped!
Re: Empirical to molecular
I believe this was one of the first examples Dr.Lavelle did in class during week 1! so if you take a look in your notes theres a step-by-step process!
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Re: Empirical to molecular
calculate what the molar mass of the empirical formula is, let's call this 'x'. The question, more often than not, will provide you with the actual molar mass of the compound ('y'). Divide 'y' by 'x', this will give you the ratio of empirical formula to molecular. Using the whole number obtained, multiply that by the coefficients of the elements from the empirical formula
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