Q.Calculate the mass per cent of different elements present in sodium sulphate ($Na_2SO_4$).

The key idea is Mass Per Cent Calculation, which involves determining the proportion of each element's mass relative to the total molar mass of the compound.

First, calculate the molar mass of sodium sulphate ($Na_2SO_4$) using the atomic masses: Na = $23 \, \text{g/mol}$, S = $32 \, \text{g/mol}$, O = $16 \, \text{g/mol}$.

Molar mass of $Na_2SO_4 = (2 \times 23) + (1 \times 32) + (4 \times 16) = 46 + 32 + 64 = 142 \, \text{g/mol}$.

Next, calculate the mass per cent for each element:

Mass per cent of Sodium ($Na$) $= \frac{\text{mass of Na}}{\text{molar mass of } Na_2SO_4} \times 100\% = \frac{46}{142} \times 100\% \approx 32.39\%$.

Mass per cent of Sulphur ($S$) $= \frac{\text{mass of S}}{\text{molar mass of } Na_2SO_4} \times 100\% = \frac{32}{142} \times 100\% \approx 22.54\%$.

Mass per cent of Oxygen ($O$) $= \frac{\text{mass of O}}{\text{molar mass of } Na_2SO_4} \times 100\% = \frac{64}{142} \times 100\% \approx 45.07\%$.

When we talk about the "mass per cent" of an element in a compound, we are essentially describing its contribution to the total mass of that compound. Imagine you have a cake, and you want to know what percentage of its total weight comes from flour, sugar, or eggs. Similarly, in chemistry, we want to know what fraction of a compound's total mass is due to sodium, sulphur, or oxygen atoms.

This concept is fundamental in chemistry for several reasons:

• Understanding Composition: It tells us the exact proportion of each element, which is crucial for understanding the compound's properties and behavior.
• Stoichiometry: It's used in quantitative analysis, for example, to determine the empirical and molecular formulas of unknown compounds or to calculate yields in chemical reactions.
• Industrial Applications: In manufacturing, knowing the mass percentage helps ensure the purity and correct composition of materials.

The approach is straightforward: we find the total mass of each element present in one mole of the compound and compare it to the total molar mass of the compound.

Here's how we calculate the mass per cent of each element in sodium sulphate ($Na_2SO_4$):

1. Identify the elements and their atomic masses.

   First, we need to know which elements are present in sodium sulphate and their respective atomic masses. The chemical formula $Na_2SO_4$ tells us we have Sodium (Na), Sulphur (S), and Oxygen (O). We'll use the standard atomic masses, often rounded for convenience in calculations unless high precision is required.

   • Atomic mass of Sodium (Na) $\approx 23 \, \text{g/mol}$
   • Atomic mass of Sulphur (S) $\approx 32 \, \text{g/mol}$
   • Atomic mass of Oxygen (O) $\approx 16 \, \text{g/mol}$

2. Determine the total mass contributed by each element in one mole of the compound.

   The subscripts in the chemical formula indicate the number of atoms of each element present in one molecule (or one formula unit) of the compound. To find the total mass contributed by each element, we multiply its atomic mass by the number of atoms of that element.

   • For Sodium (Na): There are 2 Na atoms. Total mass of Na $= 2 \times 23 \, \text{g/mol} = 46 \, \text{g/mol}$
   • For Sulphur (S): There is 1 S atom. Total mass of S $= 1 \times 32 \, \text{g/mol} = 32 \, \text{g/mol}$
   • For Oxygen (O): There are 4 O atoms. Total mass of O $= 4 \times 16 \, \text{g/mol} = 64 \, \text{g/mol}$

3. Calculate the molar mass of the compound ($Na_2SO_4$).

   The molar mass of the compound is the sum of the total masses contributed by all the elements in one mole of the compound.

   Molar mass of $Na_2SO_4$ $= (\text{Total mass of Na}) + (\text{Total mass of S}) + (\text{Total mass of O})$

   Molar mass of $Na_2SO_4$ $= 46 \, \text{g/mol} + 32 \, \text{g/mol} + 64 \, \text{g/mol}$

   Molar mass of $Na_2SO_4$ $= 142 \, \text{g/mol}$

   The molar mass of a compound is the sum of the atomic masses of all atoms in its chemical formula.

4. Calculate the mass per cent of each element.

   The mass per cent of an element is calculated by dividing the total mass contributed by that element in the compound by the molar mass of the compound, and then multiplying by 100.

   Mass per cent of an element $= \frac{\text{Total mass of the element in the compound}}{\text{Molar mass of the compound}} \times 100\%$

   • Mass per cent of Sodium (Na):

     Mass % Na $= \frac{46 \, \text{g/mol}}{142 \, \text{g/mol}} \times 100\%$

     Mass % Na $\approx 0.32394 \times 100\%$

     Mass % Na $\approx 32.39\%$

   • Mass per cent of Sulphur (S):

     Mass % S $= \frac{32 \, \text{g/mol}}{142 \, \text{g/mol}} \times 100\%$

     Mass % S $\approx 0.22535 \times 100\%$

     Mass % S $\approx 22.54\%$

   • Mass per cent of Oxygen (O):

     Mass % O $= \frac{64 \, \text{g/mol}}{142 \, \text{g/mol}} \times 100\%$

     Mass % O $\approx 0.45070 \times 100\%$

     Mass % O $\approx 45.07\%$

   Tip

   A quick check for your calculations is to sum up all the mass percentages. They should add up to approximately 100%. Any slight deviation is usually due to rounding.

   $32.39\% + 22.54\% + 45.07\% = 100.00\%$

Method: Formula Mass Percentage Method

This method calculates the mass contribution of each element relative to the total molecular mass of the compound, expressed as a percentage.

Steps

Step 1: Write the molecular formula

Sodium sulphate: $\text{Na}_2\text{SO}_4$

Step 2: Find atomic masses (from periodic table)

• Na = 23 u
• S = 32 u
• O = 16 u

Step 3: Calculate the molecular mass

Step 4: Find mass per cent of each element

• Sodium (Na): Mass of Na in one molecule = $2 \times 23 = 46$ u

• Sulphur (S): Mass of S = $32$ u

• Oxygen (O): Mass of O = $4 \times 16 = 64$ u

Final Answer

Check: $32.39 + 22.54 + 45.07 = 100\%$ ✓

Why this works

The mass per cent tells you how much of the compound’s mass comes from each element. It is independent of sample size — whether you have 1 gram or 1 kilogram of $\text{Na}_2\text{SO}_4$, the percentage composition remains the same. This is the foundation for empirical formula and stoichiometry problems in exams.

🧪 Step 1: Understand the Formula

Sodium sulphate is $Na_2SO_4$.

This means:

• 2 atoms of Sodium (Na)
• 1 atom of Sulphur (S)
• 4 atoms of Oxygen (O)

Mistake #1: Miscounting atoms

Students often forget the subscript 2 on Na or treat $SO_4$ as one atom.

✓ How to avoid: Always expand the formula: $Na_2SO_4$ = 2 Na + 1 S + 4 O.

🧮 Step 2: Use Correct Atomic Masses

From the periodic table (standard values for exams):

• Na = 23 u
• S = 32 u
• O = 16 u

Mistake #2: Using wrong atomic masses

Some use 22.99 for Na or 32.06 for S — but in most Indian exams (CBSE, ICSE, state boards), round to nearest whole number unless specified otherwise.

✓ How to avoid: Memorise the standard rounded values: H=1, C=12, N=14, O=16, Na=23, S=32, Cl=35.5, Ca=40.

➗ Step 3: Calculate Molar Mass Correctly

Molar mass of $Na_2SO_4$:

Mistake #3: Arithmetic errors

Common slip: $46 + 32 = 78$, then $78 + 64 = 142$ — but some add $46 + 32 = 78$ then forget to add 64, or add incorrectly.

✓ How to avoid: Write each step clearly. Double-check addition. Use a calculator if allowed, but show working.

📊 Step 4: Calculate Mass Percent Correctly

Mass percent formula:

So:

• Na: $\frac{46}{142} \times 100 \approx 32.39\%$
• S: $\frac{32}{142} \times 100 \approx 22.54\%$
• O: $\frac{64}{142} \times 100 \approx 45.07\%$

Mistake #4: Using wrong numerator

Some use atomic mass (23) instead of total mass (46 for Na).

✓ How to avoid: Always multiply atomic mass by the number of atoms first.

Mistake #5: Rounding too early

Rounding 46/142 to 0.32 before multiplying by 100 gives 32% instead of 32.39%.

✓ How to avoid: Keep at least 4 decimal places in division, then round final answer to 2 decimal places.

Mistake #6: Percentages don’t add to 100%

If your three percentages sum to 99.8% or 100.2%, that’s fine due to rounding. But if they sum to 95% or 105%, you made an error.

✓ How to avoid: Always check: $32.39 + 22.54 + 45.07 = 100.00\%$.

✓ Final Answer (Exam-Ready)

📝 Quick Checklist to Avoid Mistakes

• Count atoms correctly from the formula
• Use standard atomic masses (rounded)
• Multiply atomic mass × number of atoms before dividing
• Show all arithmetic steps
• Round only at the final step
• Verify percentages sum to ~100%

Final tip: Practice with $Na_2CO_3$, $Ca(OH)_2$, and $Al_2(SO_4)_3$ — the same mistakes repeat, so mastering this one saves you marks across many problems.