How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Exercises · 1.7

CuSO_4

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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The key idea is to find the mass fraction of copper in copper sulphate and multiply it by the given sample mass. From 100 g of $CuSO_4$ , you can obtain 39.8 g of copper.

Why This Approach Works

The question asks how much copper metal can be obtained from a compound. This is a classic mass-percentage problem. Copper sulphate ( $CuSO_4$ ) is a pure compound — every molecule has exactly one copper atom. So the fraction of the compound’s mass that comes from copper is fixed by the atomic masses of its elements. If you decompose the compound completely, all that copper ends up as metal. Therefore, the mass of copper you get is simply:

\text{Mass of Cu} = \left( \frac{\text{Mass of one Cu atom}}{\text{Mass of one } CuSO_4 \text{ molecule}} \right) \times \text{Mass of sample}

This ratio is the mass fraction of copper in the compound. No chemical reaction details are needed — just the formula and atomic masses.

Step-by-Step Calculation

1. Write down the atomic masses (in g/mol, which is numerically the same as the mass of one atom in amu).

From the periodic table:

Copper (Cu): 63.5 g/mol
Sulphur (S): 32 g/mol
Oxygen (O): 16 g/mol

Watch out

A common mistake is to forget that oxygen appears four times in $CuSO_4$ . Always count the subscript carefully.

2. Calculate the molar mass of $CuSO_4$ .

Add up the contributions:

Cu: $1 \times 63.5 = 63.5$ g/mol
S: $1 \times 32 = 32$ g/mol
O: $4 \times 16 = 64$ g/mol

Total:

M(CuSO_4) = 63.5 + 32 + 64 = 159.5 \text{ g/mol}

3. Find the mass fraction of copper. …

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Here are the most common mistakes students make when solving stoichiometric mass problems like this one, along with clear strategies to avoid them.

1. Forgetting to Write the Balanced Chemical Equation

The Mistake:

Students jump straight to the molar mass of $CuSO_4$ and assume the mass of copper is simply a fraction of the given mass, without considering the reaction or the fact that copper is already present in the compound.

Why It’s Wrong:

The question asks: “How much copper can be obtained?” This implies a decomposition or extraction process. Without a balanced equation, you cannot determine the mole ratio between $CuSO_4$ and $Cu$ .

How to Avoid:

Always write the relevant balanced equation first. For this problem:

CuSO_4 \rightarrow Cu + SO_4

But note: this is not balanced for charge or atoms. The correct decomposition (e.g., electrolysis or thermal) is:

2CuSO_4 \rightarrow 2Cu + 2SO_3 + O_2

However, for simple mass calculation, the key ratio is 1 mole $CuSO_4$ gives 1 mole $Cu$ . So the balanced equation is:

CuSO_4 \rightarrow Cu + S + 2O_2

But the simplest approach: 1 mole of $CuSO_4$ contains 1 mole of $Cu$ atoms. So the mole ratio is 1:1.

2. Using Incorrect Molar Masses

The Mistake:

Using atomic masses incorrectly — e.g., taking $Cu = 63.5$ , $S = 32$ , $O = 16$ but then adding them wrong, or using rounded values inconsistently.

Example of error:

$CuSO_4$ molar mass = $63.5 + 32 + (4 \times 16) = 63.5 + 32 + 64 = 159.5$ g/mol.

Some students write $159$ or $160$ and then get a slightly wrong answer.

How to Avoid:

Use standard atomic masses (given in exam or periodic table).
Write the calculation step-by-step:

M(CuSO_4) = 63.5 + 32 + 64 = 159.5 \text{ g/mol}

Bold the final molar mass before using it.

3. Confusing Mass of Compound with Mass of Element

The Mistake:

Thinking that 100 g of $CuSO_4$ contains 100 g of copper.

Why It’s Wrong:

Copper is only one part of the compound. The mass of copper is a fraction of the total mass.

How to Avoid:

Use the mass fraction method:

\text{Mass of Cu} = \frac{\text{Molar mass of Cu}}{\text{Molar mass of } CuSO_4} \times \text{mass of sample}

So:

\text{Mass of Cu} = \frac{63.5}{159.5} \times 100

4. Forgetting to Convert Grams to Moles (or Vice Versa)

The Mistake:

Directly multiplying or dividing masses without converting to moles first.

Example of error:

$100 \div 159.5 = 0.627$ (moles of $CuSO_4$ ). Then some students multiply by 63.5 but forget to multiply by the mole ratio (which is 1:1 here, so it’s fine — but in other problems, it’s not).

How to Avoid:

Always follow the mole method:

Convert given mass to moles

\text{moles of } CuSO_4 = \frac{100}{159.5}

Use mole ratio (here 1:1)

\text{moles of Cu} = \text{moles of } CuSO_4

Convert moles of Cu to mass …

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Q.How much copper can be obtained from 100 g of copper sulphate (CuSO4CuSO_4CuSO4​)?

Why This Approach Works

Step-by-Step Calculation

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Q.How much copper can be obtained from 100 g of copper sulphate ( $CuSO_4$ )?

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