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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
Answer:
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