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Question: Which of the following atoms contains the largest mass of H-atoms? A) \(5\,mol\) \({C_2}{H_2}{O_4}...

Which of the following atoms contains the largest mass of H-atoms?
A) 5mol5\,mol C2H2O4{C_2}{H_2}{O_4}
B) 1.1mol1.1\,mol C3H8O3{C_3}{H_8}{O_3}
C) 1.5mol1.5\,mol C6H8O6{C_6}{H_8}{O_6}
D) 4mol4\,mol C2H4O2{C_2}{H_4}{O_2}

Explanation

Solution

To solve this question, use the unitary method in mole concept. Calculate the mass of the Hydrogen atom in 1 mole of compound and then calculate the mass for the required number of moles using the unitary method. And at the last, simply compare the masses and you will get your answer.

Complete step-by-step solution:
Here {g refers to gram (mass)}
Step 1. For C2H2O4{C_2}{H_2}{O_4} :- 1 mole of C2H2O4{C_2}{H_2}{O_4} contains \to 2 moles of H-atom =2×1=2 = 2 \times 1 = 2 g of H-atom.
So, 5 moles of C2H2O4{C_2}{H_2}{O_4} contains \to 5×25 \times 2 g =10 = 10 g of H-atoms.

Step 2. For C3H8O3{C_3}{H_8}{O_3} :- 1 mole of C3H8O3{C_3}{H_8}{O_3} contains \to 8 moles of H-atom =8×1=8 = 8 \times 1 = 8 g of H-atom.
So, 1.1 moles of C3H8O3{C_3}{H_8}{O_3} contains \to 1.1×81.1 \times 8 g =8.8 = 8.8 g of H-atoms.

Step 3. For C6H8O6{C_6}{H_8}{O_6} :- 1 mole of C6H8O6{C_6}{H_8}{O_6} contains \to 8 moles of H-atom =8×1=8 = 8 \times 1 = 8 g of H-atom.
So, 1.5 moles of C6H8O6{C_6}{H_8}{O_6} contains \to 1.5×81.5 \times 8 g =15 = 15 g of H-atom.

Step 4. For C2H4O2{C_2}{H_4}{O_2} :- 1 mole of C2H4O2{C_2}{H_4}{O_2} contains \to 4 moles of H-atom =4×1=4 = 4 \times 1 = 4 g of H-atom.
So, 4 moles of C2H4O2{C_2}{H_4}{O_2} contains \to 4×4=164 \times 4 = 16g of H-atom.
So, by comparing all the masses we can conclude that 4 moles of C2H4O2{C_2}{H_4}{O_2} will contain the greatest mass of H-atoms.

Hence, option D is the correct answer.

Note: A substance's molar mass is calculated by multiplying its relative atomic mass by the molar mass constant (1 g/mol). The molar mass constant can be used to convert mass to moles. By multiplying a given mass by the molar mass, the amount of moles of the substance can be calculated.