# Question 36c25

Nov 3, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to use the mole ratio that exists between lithium hydrogen carbonate, ${\text{LiHCO}}_{3}$, and lithium carbonate, ${\text{Li"_2"CO}}_{3}$, to determine how much of the former compound the original sample contained.

Take a look at the balanced chemical equation for this decomposition reaction. Carbonic acid, ${\text{H"_2"CO}}_{3}$, will actually decompose to form water and carbon dioxide.

color(red)(2)"LiHCO"_text(3(s]) -> "Li"_2"CO"_text(3(s]) + overbrace("H"_2"O"_text((l]) + "CO"_text(2(g]))^(color(blue)("H"_2"CO"_3))

Notice that being a gas, the carbon dioxide is evolved; moreover, heating the sample will dry off the water, so you can say that the mass of the remaining sample will be equal to the mass of lithium carbonate.

Use lithium carbonate's molar mass to determine how many moles were produced by the reaction

2.501color(red)(cancel(color(black)("g"))) * ("1 mole Li"_2"CO"_3)/(73.891color(red)(cancel(color(black)("g")))) = "0.033847 moles Li"_2"CO"_3

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between lithium hydrogen carbonate and lithium carbonate. This means that the reaction will produce $1$ mole of the latter for every $\textcolor{red}{2}$ moles of the former.

SInce you know how many moles of lithium carbonate were produced, you can say that the original sample contained

0.033847color(red)(cancel(color(black)("moles Li"_2"CO"_3))) * (color(red)(2)" moles LiHCO"_3)/(1color(red)(cancel(color(black)("mole Li"_2"CO"_3)))) = "0.067694 moles LiHCO"_3

Now use lithium hydrogen carbonatye's molar mass to find out how many grams would contain this many moles

0.067694color(red)(cancel(color(black)("moles"))) * "67.958 g"/(1color(red)(cancel(color(black)("mole")))) = "4.600 g LIHCO"_3#

As you can see, this is a problem because your original sample had a mass of $\text{4.076 g}$, so it couldn't have possibly contained $\text{4.600 g}$ of lithium hydrogen carbonate.

So, make sure that you used the correct values for the initial amss and/or the remaining mass. My guess is that the mass of lithium carbonate is somewhere between $\text{1.5 g}$ and $\text{2.0 g}$.

Once you get the correct values, redo the calculations with what you have.