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Chapter 5: Problem 48

A diamond contains \(5.0 \times 10^{21}\) atoms of carbon. What amount (moles) of carbon and what mass (grams) of carbon are in this diamond?

Short Answer

Expert verified
The diamond contains \( 8.304 \times 10^{-3} \) moles of carbon and \( 9.970 \times 10^{-2} \) grams of carbon.

Step by step solution

01

Recall Avogadro's number and the molar mass of carbon

First, we need to recall Avogadro's number, which is the number of particles (such as atoms) per mole of a substance. Avogadro's number is approximately \(6.022 \times 10^{23} \) particles per mole. We also need to know the molar mass of carbon, which is approximately \(12.01\) grams per mole.
02

Convert atoms of carbon to moles of carbon

Given that the diamond contains \(5.0 \times 10^{21}\) atoms of carbon, we can use Avogadro's number to calculate the number of moles of carbon in the diamond: Moles of carbon = \( \frac{atoms \ of \ carbon}{Avogadro's \ number} \) Moles of carbon = \( \frac{5.0 \times 10^{21}}{6.022 \times 10^{23}} \) Moles of carbon = \(8.304 \times 10^{-3} \) moles
03

Convert moles of carbon to grams of carbon

Now we can use the molar mass of carbon to convert the moles of carbon calculated in step 2 to grams of carbon: Mass of carbon = Moles of carbon 脳 Molar mass of carbon Mass of carbon = \( 8.304 \times 10^{-3} \ moles \times 12.01 \ g/mol \) Mass of carbon = \( 9.970 \times 10^{-2} \) grams
04

Write the final answer

Thus, the diamond contains \( 8.304 \times 10^{-3} \) moles of carbon and \( 9.970 \times 10^{-2} \) grams of carbon.

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Most popular questions from this chapter

Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ (Cu does not react with HCl.) When 0.5065 g of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{g}\) \(\mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

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