Chemistry Q51 – JEE-MAIN 2026

Home Chemistry Basic ChemistryQuestion #64

An oxide of iron contains

69.9 %

iron, its empirical formula, is: (Given : Molar mass of Fe and O are

56

and

16

g mol

^{- 1}

respectively.)

Clue 1 of 3: Percentage Composition

Assume a convenient total mass (e.g., 100 g) to convert percentages into masses of individual elements.

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Step 1: Calculate mass of each element✦ Active

Assume $100$ g of the iron oxide. The mass of iron (Fe) is $69.9$ g. The mass of oxygen (O) is the total mass minus the mass of iron.

Mass of Fe = 69.9 g

Mass of O = 100 g - 69.9 g = 30.1 g

Step 2: Calculate moles of each element○ Expand

Using the given molar masses ( $56$ g/mol for Fe and $16$ g/mol for O), calculate the moles of each element.

Moles of Fe = \frac{69.9 g}{56 g/mol} \approx 1.248 mol

Moles of O = \frac{30.1 g}{16 g/mol} \approx 1.881 mol

Step 3: Determine the simplest whole number ratio and empirical formula○ Expand

Divide the moles of each element by the smallest number of moles to find the ratio. Then, multiply by a small integer to get whole numbers for the empirical formula.

Ratio of Fe = \frac{1.248}{1.248} = 1

Ratio of O = \frac{1.881}{1.248} \approx 1.507

To get whole numbers, multiply by 2:

Fe : 1 \times 2 = 2

O : 1.507 \times 2 \approx 3.014 \approx 3

The empirical formula is F e_{2} O_{3}

💡 Teacher's Secret Hint

Remember to round to the nearest whole number only after multiplying to eliminate fractions, if necessary.

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Chemistry Question 51 – JEE-MAIN 2026

Clue 1 of 3: Percentage Composition

🧠 Full Solution Path