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31 October, 19:06

A compound is found to contain 9.227 % boron and 90.77 % chlorine by mass. What is the empirical formula for this compound?

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  1. 31 October, 19:22
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    Find the moles of Boron and of Chlorine:

    Assume there are 100 g of the compound in all

    Boron: 9.224% - ->.09224 (convert percent to decimals)

    .09224*100g = 9.224g Boron (theoretical amt)

    9.224g * 1mol/10.81g (molar mass Boron) =.8533 moles Boron
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