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