Question

Question > calculate the percent composition of each element in the following compounds

Answer 1

Answer:

The percent composition of each element are :

1. Th(SO4)2 :

Th = 54.704%  ,  S =  15.119%  ,    O = 30.1759 %

2.Ca3(PO4)2

Ca = 38.76%    ,   P = 19.97% ,       O = 41.265%

3.C6H12O6

C = 40.0%        ,   H = 6.72% ,         O = 53.29%

Explanation:

Formula  calculate the percent composition

$% composition =\frac{mass\ of\ given\ element}{Molar\ mass}$

Molar mass of Th(SO4)2 = Mass of Th + 2(mass of S) + 8(mass of O)

= 232.03 + 2(32.06) + 8(16)

=232.03 +64.12+128

= 424.16

$% Th=\frac{mass\ of\ Th}{424.16}$

$=\frac{232.03}{424.16}$

= 54.704%

$% S=\frac{2(mass\ of\ S)}{424.16}$

$=\frac{64.12}{424.16}$

= 15.119%

$% O=\frac{8(mass\ of\ O)}{424.16}$

$= \frac{128}{424.16}$

= 30.17%

2.Ca3(PO4)2

Molar mass of Ca3(PO4)2 = 3(Mass of Ca) + 2(mass of P) + 8(mass of O)

= 3(40) + 2(30 .9) + 8(16)

= 120 + 61.8 + 128

=310.176

$% Ca =\frac{3(mass\ of\ Ca)}{310.176}$

$=\frac{120}{424.16}$

38.76%

$% P=\frac{2(mass\ of\ P)}{310.176}$

$=\frac{61.8}{310.176}$

= 19.97%

$% O=\frac{8(mass\ of\ O)}{310.176}$

$=\frac{128}{424.16}$

= 41.26 %

3.

C6H12O6

Molar mass of C6H12O6  = 6(Mass of C) + 12(mass of O) + 6(mass of O)

= 6(12) + 12(1) + 6(15.99)

=180

$% O=\frac{6(mass\ of\ O)}{180}$

$=\frac{95.94}{180}$

= 53.28%

$% C=\frac{6(mass\ of\ C)}{180}$

$% C=\frac{72}{180}$

= 40%

$% H=\frac{12(mass\ of\ C)}{180}$

$=\frac{(12}{180}$

= 6.713%