Question

How many micrograms of hydrogen are in 8.99 x 10-9 decigrams of lithium hydrogen carbonate octahydrate?

Answer 1

Llithium hydrogen carbonate octahydrate is $LiHCO_3 \cdot 8H_2O$.

$m[\mu g]=m[dg] \cdot 10^4$

$m(LiHCO_3 \cdot 8H_2O)=8.99 \cdot 10^{-9} \cdot 10^{4}=8.99 \cdot 10^{-5} \; \mu g$

$m(H)= \frac{m(LiHCO_3 \cdot 8H_2O) \cdot M(H) \cdot n(H)}{M(LiHCO_3 \cdot 8H_2O)}$

$n(H)=1+8 \cdot 2=17$

$M(H)=1.008 \; g/mol$

$M(LiHCO_3 \cdot 8H_2O)=1 \cdot 6.941 +1 \cdot 1.008 + 1 \cdot 12.011 + 3 \cdot 15.999 + 8 \cdot (2 \cdot 1.008 + 1 \cdot 15.999)=212.077 \; g/mol$

$m(H)= \frac{8.99 \cdot 10^{-5} \cdot 1.008 \cdot 17}{212.077} \approx 7.26 \cdot 10^{-6} \; \mu g$