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Question

Question: How do you calculate the moles of a substance?...

How do you calculate the moles of a substance?

Explanation

Solution

In order to calculate the moles of a substance, you need to know the mass of the substance and its molar mass. Molar mass is the atomic weight in grams/mol. Utilize the atomic formula to locate the molar mass; to acquire the quantity of moles, partition the mass of compound by the molar mass of the compound communicated in grams

Complete step by step answer:
To calculate moles of copper sulfate, Ca(OH)2Ca(OH)_2 in 250.0g Ca(OH)2Ca(OH)_2?
Molar Mass of CuSO4CuSO_4
Subscript x molar mass =
1  40.0781{\text{ }}*{\text{ }}40.078 g/mol Ca = 40.07840.078 g/mol Ca
1  1.007g1{\text{ }}*{\text{ }}1.007g /mol H = 1.007g1.007g /mol H
2 x 15.9992{\text{ }}x{\text{ }}15.999 g/mol O = 31.99831.998 g/mol O
Total Molar mass of Ca(OH)2:73.083g/molCa(OH)_2: 73.083g/mol
1mol Ca(OH)2Ca(OH)_2 = 73.083g Ca(OH)2Ca(OH)_2 , which gives us two conversion factors:
1mol Ca(OH)273.083g Ca(OH)2\dfrac{{1{\text{mol Ca}}{{\left( {OH} \right)}_2}}}{{73.083{\text{g Ca}}{{\left( {OH} \right)}_2}}} and 73.083gCa(OH)21mol Ca(OH)2\dfrac{{73.083{\text{gCa}}{{\left( {OH} \right)}_2}}}{{1{\text{mol Ca}}{{\left( {OH} \right)}_2}}}

Now multiply the known mass x the conversion factor with moles on top and grams on bottom. This will cancel the grams and leave the moles.

250.0g Ca(OH)2250.0g{\text{ }}Ca{\left( {OH} \right)_2} x1molCa(OH)273.083gCa(OH)2\dfrac{{1mol{\text{Ca}}{{\left( {OH} \right)}_2}}}{{73.083g{\text{Ca}}{{\left( {OH} \right)}_2}}} =   3.42\;3.42 mol Ca(OH)2mol{\text{ }}Ca{\left( {OH} \right)_2}

Note:
The mole is spoken to by Avogadro's number, which is 6.022×10236.022 \times {10^{23}} particles or atoms per mol.
Mole : The measure of substance of a framework that contains the same number of rudimentary elements as there are molecules in 12 g12{\text{ }}g of carbon12carbon - 12 .
In order to calculate the molar mass of a substance you should take the approx value.
Example: -
Take molar mass of CaCa =4040 g/mol in the place of 40.07840.078 g/mol
Take molar mass of HH =1 g/mol in the place of 1.007 g/mol
Take molar mass of OO = 16 g/mol in the place of 16 g/mol