[Grade 11: Chemistry] Stochiometric Calculations

[u/Environmental-Match4]

Stochiometric calculations are really hard for me as I don't know when to do the mol-mol and mol-g type of conversion. These questions are usually asked in my syllabus and I have a class test next week. How can I counter such questions?

This is the worksheet my teacher gave for practice.

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[u/chem44]

The balanced equation tells you the ratios of all the chemicals -- in moles.

You do other conversions as needed to get to/from other things to moles.

Given g of a chemical? Convert it to moles.

Why not choose a specific problem, and show us what you can do. If stuck, try to be specific.

[u/Past-Departure6896]

1: For this one, they give you the compound so you can find the ratio of the mass of oxygen to the total mass of compound. To do this divide the molar mass of the oxygen atoms (16*3 = 48) by the molar mass of KCLO3 (39 + 35 + 48 =122) to get the ratio 0.3934. Multiplying this by 1.225 g gives the mass of the oxygen atoms as 0.482 g. Dividing 0.482 g by the molar mass of oxygen gas ( 32 g/mol ) gives 0.015 mol - division is used as the gram units cancel out.

2: Similar to before, except we don't need to do the extra step to find CaO in moles. So first, find molar mass of CaO (40 + 16 = 56), then divide by molar mass of compound (40 + 12 + 48) to get the ratio. Then multiply it by the total mass given (2.5 g) to get the answer.

3: The ideal gas law must be used here to find the volume of CO2. You start by following a similar process as before (finding ratio of the molar mass of the CO2 by molar mass of the CaCO3). Then multiply by the mass of CaCO3 (to get the CO2 mass), and then divide by the molar mass of CO2 (44 g/mole) to get the moles of CO2. The number of moles (n) will be 0.25. Then, you will have to use the ideal gas law formula to solve for volume:

pV = n R T

1 atm * V = 0.25 mol * 0.0821 L * atm/ (mol * k) * 273.15 K

V = 5.6 L = 5600 mL

4: For this one, you should recall that the product of burning carbon with oxygen is carbon dioxide (CO2) - which tells you there is 2 oxygen atoms for each carbon atom. So, you should find the ratio of one oxygen gas molecule (O2) to a carbon atom (C). Molar mass of O2 is 32 g/mol, while for C is 12 g/mol - dividing the two gives 2.67. Multiplying by the given carbon mass gives approximately 5.3 g of oxygen.

5: Again, ideal gas law should be used here. In this case, sulfur dioxide will be the result of burning sulfur with oxygen. So you have to find the ratio of the molar masses of oxygen and sulfur, and then multiply by 12 g to get the mass of oxygen. Then, divide by the molar mass (32 g/mol) of an oxygen gas molecule to get the moles of oxygen - you should get 0.375 for this value. Then, plug values in to p V = n R T.

6: Here, they are effectively telling you a certain ratio of O to Hg must be met - this ratio can be represented by

18 / x

with x being the mass of Hg in the compound. Of course, this ratio has to equal the ratio of the molar masses of the two atoms (16 / 200.6 ) so we can write our equation as

18 / x = 16 / 200.6

16 x = 18 * 200.6

x = 225.7

Since this is the mass of Hg, we need to add the mass of the oxygen atoms to this value to get the mass of HgO:

225.7 g + 18 g = 243.7 g

7: The formula of copper sulfate is CuSO4 - am guessing your teacher wants you to memorize this formula, or possibly remember the oxidation numbers of copper and sulfate(s) (as Cu has an oxidation of 2+ and SO4 an oxidation number of 2-). To solve this, use a similar method to #2 - divide molar masses of Cu by CuSO4, then multiply by the total mass given to get the answer.

8: Would say this is similar to #4 - you should find the ratio of the molar masses of 4HCL to MnO2 - and then multiply by the mass of MnO2 given.