The class of ternary compounds called carboranes (containing the elements carbon

Question

The class of ternary compounds calledcarboranes (containing the elements carbon, boron andhydrogen) have played a prominent role in many areas of fundamentalchemical research. In particular, they have provided manyinteresting examples for testing theories on chemical bonding andstructure.
A particular carborane has the following mass percentages: 21.30%Cand 11.62%H.
What is the empirical formula of thiscompound?
(If, for example, the compound was CB3H5,then one must input CB3H5 --- no spaces, nocommas, uppercase, and the order in which the element symbolsappear is C 1st, B 2nd, H 3rd.) I'll rate lifesavior! The class of ternary compounds calledcarboranes (containing the elements carbon, boron andhydrogen) have played a prominent role in many areas of fundamentalchemical research. In particular, they have provided manyinteresting examples for testing theories on chemical bonding andstructure.
A particular carborane has the following mass percentages: 21.30%Cand 11.62%H.
What is the empirical formula of thiscompound?
(If, for example, the compound was CB3H5,then one must input CB3H5 --- no spaces, nocommas, uppercase, and the order in which the element symbolsappear is C 1st, B 2nd, H 3rd.) I'll rate lifesavior!

Explanation / Answer

So you can solve this by percent masses: If you know the percentages of carbon and hydrogen you can subtractthem from a 100 to get the percent of boron: 100-21.3-11.62=67.08 So then you can pretend you have a 100 g of your compound and thateach corresponding percent is the grams: 21.3 g C 11.62g H 67.08g B To figure out an empirical formula you need moles so you mustconvert g to moles which you can do by viewing the periodic table.The molar mass of an elements gives you how many grams are in amole of the substance: 21.3g C x 1 mol/12.01 g= 1.77 mol C 11.62g H x 1mol/1.008 g= 11.53 mol H 67.08g B x 1mol/10.81 g = 6.205 mol B So since C has the smallest mol value you divide all of them byit 1.77/1.77= 1 C 11.53/1.77= 6.5 H 6.205/1.77= 3.5 B Those would be your numbers for the moles of each, however since Hand B have ".5" you must double all of the numbers since you can'thave half a mol in an empirical formula So you come up with: C2H13B7

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