Question

determine the quantity of hydrogen atoms in 0.1488 g of phosphoric acid, h₃po₄.
(starting amount section, then a multiplication with factors, and a set of buttons with values like 98.00, 3, 3.657×10²¹, 4, 2.743×10²¹, 0.1488, 9.144×10²⁰, 3.048×10²⁰, 1, 52.02, 6.022×10²³, 1.01 and units like mol h, g h, g h₃po₄/mol, molecules h₃po₄, mol h₃po₄, g h₃po₄, atoms h)

Explanation:

Step1: Start with given mass

$0.1488\ \text{g}\ \text{H}_3\text{PO}_4$

Step2: Convert to moles of $\text{H}_3\text{PO}_4$

Multiply by $\frac{1\ \text{mol}\ \text{H}_3\text{PO}_4}{98.00\ \text{g}\ \text{H}_3\text{PO}_4}$
Expression: $0.1488\ \text{g}\ \text{H}_3\text{PO}_4 \times \frac{1\ \text{mol}\ \text{H}_3\text{PO}_4}{98.00\ \text{g}\ \text{H}_3\text{PO}_4}$

Step3: Convert to moles of H

Multiply by $\frac{3\ \text{mol}\ \text{H}}{1\ \text{mol}\ \text{H}_3\text{PO}_4}$
Expression: $0.1488 \times \frac{1}{98.00} \times \frac{3\ \text{mol}\ \text{H}}{1\ \text{mol}\ \text{H}_3\text{PO}_4}$

Step4: Convert to atoms of H

Multiply by $\frac{6.022 \times 10^{23}\ \text{atoms H}}{1\ \text{mol H}}$
Expression: $0.1488 \times \frac{1}{98.00} \times 3 \times 6.022 \times 10^{23}$

Step5: Calculate final value

$\frac{0.1488 \times 3 \times 6.022 \times 10^{23}}{98.00} = 2.743 \times 10^{21}$

Answer:

$2.743 \times 10^{21}$ atoms H