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Posted by **Amanda** on Tuesday, September 29, 2009 at 10:50pm.

CaCO3(s) → CaO(s) + CO2(g)

What is the percentage of silica in the sandstone if 19.2 mg of the rock yields 4.21 mg of carbon dioxide?

- Chemistry -
**JOHNNY**, Tuesday, September 29, 2009 at 10:52pmFor my explanation I will use "What is the percentage of silica in the sandstone if 18.0 GRAMS of the rock yields 4.4 GRAMS of carbon dioxide?" as the question. Note that you will have to convert mg to g.

First, balance the equation and write it down. This equation is already balanced, so 1 CaCO3--------> 1 CaO + 1 CO2.

Next, ask yourself "What information do I need in order to solve the problem?" We need 2 of the following 3 values:

1) mass of sandstone, we know is 18g

2) mass of silica in the sandstone

3) mass of CaCO3 originally in the sandstone

We are only given the mass of the sandstone, therefore, there must be a way to determine one of the other two values from the information we were presented. In our case, we will deduce the amount of CaCO3 originally present using the amount of CO2 produced.

In any balanced equation, the coefficients of the substances directly relate to the molar ratio between any substances involved. That is, we can say "One mole of CaCO3, upon decomposition, will produce exactly one mole of CO2," because each has a coefficient of 1 in the balanced equation. These statements can be made in "reverse" as well. If we get one mole of CO2 produced, we know that there was one mole of CaCO3 present originally.

Once you realize that a problem is going to be dealing with moles, grams, etc., write down each substance involved, along with its molecular weight and how much of the substance you have.

CaCO3----------mol. wt. 100-------------______ grams

CaO--------------mol. wt. 56--------------______ grams

CO2--------------mol. wt. 44----------------- 4.4 grams

SiO2-------------mol. wt. 60.1--------------_____ grams

Sandstone-------not known------------------18.0 grams

If we know how many moles of CO2 were produced, then we know how many moles of CaCO3 were present initially.

We are given CO2 in grams, to convert to moles use the following:

...............[1 mol CO2 ]

4.4g CO2 [-----------------] = 0.1 moles of CO2

...............[ 44g CO2 ]

Since the coefficients are the same, we know that this translates to 0.1 moles of CaCO3 initially present. To convert to grams:

...........................[ 100g CaCO3 ]

0.1 moles CaCO3 [---------------------] = 10g CaCO3

...........................[1 mol CaCO3 ]

If you start with moles, then put moles on the bottom and molecular weight on top. The molecular weight of CaCO3 is 100g/mol. The moles cancel out leaving grams. If you start with grams, put grams on the bottom and moles on top.

Put these values in the table as you go.

CaCO3----------mol. wt. 100-------------__10__ grams

CaO--------------mol. wt. 56--------------______ grams

CO2--------------mol. wt. 44----------------- 4.4 grams

SiO2-------------mol. wt. 60.1--------------_____ grams

Sandstone-------not known------------------18.0 grams

We now know 2 of the 3 original values that were needed. Mass of sandstone = 18g, and mass of CaCO3 = 10g.

Since the sandstone is composed of only two substances, CaCO3 and SiO2, figuring one percentage will automatically give us the other. Percent composition =

Mass part

------------------------ =

Mass total

10g CaCO3

--------------------------- = 56% CaCO3, leaving 44% silica.

18g sandstone

Now work through the problem step by step with your values. Don't forget to convert the mg to grams.

- Chemistry -
**DrBob222**, Tuesday, September 29, 2009 at 10:58pmConvert 4.21 mg CO2 to moles. 1 mole CO2 = 1 mole CaCO3, then convert moles CaCO3 to grams CaCO3.

%CaCO3 = (g CaCO3/0.0192 g sample)*100 = ??

% SiO2 = 100%-%CaCO3.

Post your work if you have any trouble.*By the way, there is a MUCH easier way to do this but it isn't taught anymore since everyone has gone to SI units.*

The easy way to convert CO2 to CaCO3 is

4.21 mg CO2 x (molar mass CaCO3/molar mass CO2) = 4.21 x (100/44) = ?? mg CaCO3. That molar mass CaCO3/molar mass CO2 is called the "gravimetric factor" but it isn't in any of the modern texts.

- Chemistry -
**Amanda**, Tuesday, September 29, 2009 at 11:08pmThanks so much. These examples really help me out.