Question

calculate the no of atoms of each element in 0.18 g of glucose

Answer 1

Dear Student,
GIVEN : mass of C₆H₁₂O₆ = 0.18 gm
molar mass of C₆H₁₂O₆ = 180 gm

1) C
% C = in glucose = $\frac{72}{180}\times 100=40\%$
So weight of C in 0.18 gm = $\frac{40}{100}\times 0.18$ =0.72 gm
12 g = $6.023\times {10}^{23}$ atoms
so 0.72 gm = 3.6 $\times {10}^{22}$ atoms
so number of atoms of C=$3.6\times {10}^{22}$
2) H
%H = $\frac{12}{180}\times 100$= 6.66%
so weight of H in 0.18 gm glucose = $\frac{6.66}{100}\times 0.18$=0.012 gm
1 gm H = 6.023 $\times {10}^{23}$
0.012 gm = 7.2 $\times {10}^{21}$ atoms
3) O
%O =$\frac{96}{180}\times 100=$ 53.3%
so weight in 0.18 gm = $\frac{53.3}{100}\times 0.18=0.095$gm
16 gm O = $6.023\times {10}^{23}$ atoms
0.095 gm O = $3.57\times {10}^{21}$ atoms
Regards!