Chemistry help beware: moles inside

[FancyNut]

Ok chemistry is my WORST subject... so go ahead and laugh first.

ahem.

I have this question that I can't seem to figure out the answer for... or more likely just figure out how to start... here it is:

A 2.5783 g mixture containing malachite and an inert ingredient lost 0.3121 g of H_2O and CO_2 (... i know i know) when decomposed by heating to constant mass...

-- How many moles of malachite in the mixture produced this mass of CO_2 and H_2O?

The thing is I'd think I know how to do it with an equation telling me ratios. With this it's just.. I don't know what to do since I don't know the ratio of malachite moles in that equation to H_2O and CO_2...

:(

 

[member 553083]

Well, the first step is to find the masses of malachite and the inert ingredient. You can do that by subtracting the mass of H_2O and CO_2 from the original mixture. Then you need to determine the moles of each substance in the mixture. The easiest way to do this is by using the molar mass of each substance. Once you have the moles of each substance, you can calculate the ratio of malachite to H_2O and CO_2. Finally, you can use this ratio to calculate the moles of malachite that produced the mass of H_2O and CO_2.

 

[wais]

First of all, don't worry, you're not alone in struggling with chemistry! It can be a challenging subject for many students. But don't let that discourage you, with some practice and help, you can definitely improve your understanding of it.

Now, onto your question. This type of problem is known as a stoichiometry problem, where you need to use ratios and conversions to find the answer. In this case, we are looking for the number of moles of malachite in the mixture, given the mass of CO_2 and H_2O produced.

The key here is to use the molar mass of each compound. The molar mass of CO_2 is 44.01 g/mol and the molar mass of H_2O is 18.02 g/mol. We can use these values to convert the mass of CO_2 and H_2O lost to moles.

First, let's convert the mass of CO_2 lost to moles. We know that 0.3121 g of CO_2 was produced, so we can use the following calculation:

0.3121 g CO_2 x (1 mol CO_2/44.01 g CO_2) = 0.0071 mol CO_2

Similarly, we can convert the mass of H_2O lost to moles:

0.3121 g H_2O x (1 mol H_2O/18.02 g H_2O) = 0.0173 mol H_2O

Now, we need to find the mole ratio of malachite to CO_2 and H_2O in the reaction. To do this, we need to look at the balanced chemical equation for the decomposition of malachite:

2CuCO_3•Cu(OH)_2 → 3CO_2 + H_2O + 2CuO

From this equation, we can see that for every 2 moles of malachite, we get 3 moles of CO_2 and 1 mole of H_2O. This means that the mole ratio of malachite to CO_2 is 2:3 and the mole ratio of malachite to H_2O is 2:1.

Now, we can use this mole ratio to find the number of moles of malachite in the mixture:

0.0071 mol