Boron forms an extensive series of compounds with hydrogen, all with the general

Question

Boron forms an extensive series of compounds with hydrogen, all with the general formula BxHy. If 0.148 g of BxHy gives 0.422 g of B2O3 when burned in excess of O2, what is the empirical formula of BxHy?

The answer is B5H7, but I want to see work on how to get there. Boron forms an extensive series of compounds with hydrogen, all with the general formula BxHy. BxHy(s) + excess O2(g) ---> x / 2 B2O3(s) + y/2 H2O (g) If 0.148 g of BxHy gives 0.422 g of B2O3 when burned in excess of O2, what is the empirical formula of BxHy?

Explanation / Answer

grams of B in 0.422 g of B2O3:

0.422 g times (21.622 / 69.619) = 0.131 g

grams of H in 0.148 g of BxHy:

0.148 - 0.131 = 0.017 g

moles B in 0.131 g ---> 0.01212963
moles H in 0.017 g ---> 0.01686508

look for smallest whole-number ratio:

B ---> 0.01212963 /0.01212963 = 1
H ---> 0.01686508 / 0.01212963 = 1.4

multiply by 10 gives 10 and 14.

remove common factor of 2 gives 5 and 7

B5H7

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