Question

how many atoms of hydrogen are present in a sample of c₆h₈o₆ that has 29.5 g of carbon?
29.5 g c | 1 mol c | 8 mol h | 6.02 x 10²³ atoms h

12.01 g c

6 mol c

1 mol h

? x 10^? atoms h
enter the coefficient in the green box and the exponent in the yellow box. report your answer using significant figures.

Explanation:

Step1: Calculate moles of C

First, convert grams of C to moles. The molar mass of C is 12.01 g/mol. So, moles of C = $\frac{29.5\ g}{12.01\ g/mol}$.

Step2: Relate moles of C to moles of H

In $\ce{C6H8O6}$, the mole ratio of H to C is $\frac{8\ mol\ H}{6\ mol\ C}$. Multiply moles of C by this ratio to get moles of H.

Step3: Convert moles of H to atoms of H

Use Avogadro's number ($6.02\times10^{23}$ atoms/mol) to convert moles of H to atoms. Multiply moles of H by $6.02\times10^{23}$ atoms/mol.

Putting it all together:
\[ \begin{align*} \text{Atoms of H} &= 29.5\ g\ C\times\frac{1\ mol\ C}{12.01\ g\ C}\times\frac{8\ mol\ H}{6\ mol\ C}\times\frac{6.02\times10^{23}\ atoms\ H}{1\ mol\ H}\\ &= 29.5\times\frac{1}{12.01}\times\frac{8}{6}\times6.02\times10^{23}\\ &\approx 29.5\times0.08326\times1.333\times6.02\times10^{23}\\ &\approx 29.5\times0.111\times6.02\times10^{23}\\ &\approx 3.2745\times6.02\times10^{23}\\ &\approx 19.71\times10^{23}\\ &\approx 1.97\times10^{24} \end{align*} \]

Wait, let's recalculate more accurately:

First, $\frac{29.5}{12.01} \approx 2.456\ mol\ C$

Then, $\frac{8}{6} \times 2.456\ mol\ C = \frac{4}{3} \times 2.456 \approx 3.275\ mol\ H$

Then, $3.275\ mol\ H \times 6.02\times10^{23}\ atoms/mol \approx 3.275\times6.02\times10^{23} \approx 19.7155\times10^{23} = 1.97\times10^{24}$ (adjusting significant figures: 29.5 has 3 sig figs, 12.01 has 4, 8 and 6 are exact, 6.02e23 has 3. So the result should have 3 sig figs.)

Wait, let's do the calculation step by step with the given conversion factors:

$29.5\ g\ C \times \frac{1\ mol\ C}{12.01\ g\ C} = \frac{29.5}{12.01} \approx 2.456\ mol\ C$

$2.456\ mol\ C \times \frac{8\ mol\ H}{6\ mol\ C} = 2.456\times\frac{8}{6} \approx 3.275\ mol\ H$

$3.275\ mol\ H \times 6.02\times10^{23}\ atoms\ H/mol\ H = 3.275\times6.02\times10^{23} \approx 19.7155\times10^{23} = 1.97\times10^{24}$ (when rounded to 3 significant figures, since 29.5 has 3, 12.01 has 4, but the least number of significant figures in the given data for multiplication/division is 3 from 29.5 and 6.02e23 (3 sig figs). Wait, 6.02e23 has 3, 29.5 has 3, 12.01 has 4, 8 and 6 are exact. So the result should have 3 sig figs.

Wait, let's check the calculation again:

$29.5 \div 12.01 = 2.456286$

$2.456286 \times 8 = 19.65029$

$19.65029 \div 6 = 3.275048$

$3.275048 \times 6.02e23 = 3.275048\times6.02\times10^{23}$

$3.275048\times6.02 = 19.71579$

So $19.71579\times10^{23} = 1.971579\times10^{24} \approx 1.97\times10^{24}$ (3 sig figs)

Wait, but maybe I made a mistake in sig figs. Let's see: 29.5 (3), 12.01 (4), 8 (exact), 6 (exact), 6.02e23 (3). So the limiting is 3 sig figs. So the coefficient is 1.97 and the exponent is 24? Wait, no, wait:

Wait, $19.71579\times10^{23} = 1.971579\times10^{24}$, so the coefficient is approximately 1.97 (3 sig figs) and the exponent is 24.

Wait, but let's do the calculation with more precision:

$29.5 \times 8 \times 6.02 \div (12.01 \times 6)$

$29.5\times8 = 236$

$236\times6.02 = 236\times6 + 236\times0.02 = 1416 + 4.72 = 1420.72$

$12.01\times6 = 72.06$

$1420.72 \div 72.06 \approx 19.715$

So $19.715\times10^{23} = 1.9715\times10^{24} \approx 1.97\times10^{24}$ (3 sig figs)

So the coefficient is 1.97 and the exponent is 24.

Wait, but let's check the original problem's conversion factors. The user provided:

29.5 g C * (1 mol C / 12.01 g C) * (8 mol H / 6 mol C) * (6.02 x 10^23 atoms H / 1 mol H)

So multiplying all the numbers:

29.5 * 1 * 8 * 6.02 x 10^23 / (12.01 * 6 * 1)

Calculate numerator: 29.5 * 8 * 6.02 = 29.5 * 48.16 = 1420.72

Denominator: 12.01 * 6 = 72.06

1420.72 / 72.06 ≈ 19.715

So 19.715 x 10^23 = 1.9715 x 10^24, which rounds to 1.97 x 10^24 (3 sig figs). So the coefficient is 1.97 and the exponent is 24.

Wait, but maybe the problem expects the coefficient to be 1.97 and exponent 24? Let me confirm.

Alternatively, maybe I made a mistake in the exponent. Wait, 19.715 x 10^23 is 1.9715 x 10^24, yes. So the coefficient is 1.97 (or 1.97 when rounded to 3 sig figs) and the exponent is 24.

So the green box (coefficient) is 1.97 and the yellow box (exponent) is 24.

Answer:

Green box: 1.97, Yellow box: 24