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High School Chemistry : Using Avogadro's Number

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Example Questions

Example Question #1 : Using Avogadro's Number

How many atoms are in $\small 50g$ of calcium?

Possible Answers:

$3.0*10^{25}atoms$

$6.0*10^{23}atoms$

$5.0*10^{24} atoms$

$7.5*10^{23} atoms$

Correct answer:

$7.5*10^{23} atoms$

Explanation:

In order to determine how many atoms are in this sample, we need to convert this sample into moles. Calcium has a molar mass of $\small 40.08$ grams per mole.

$50g\ Ca*\frac{1mol\ Ca}{40.08g\ Ca}=1.25mol\ Ca$

Avogadro's number tells us that there $6.022*10^{23}$ atoms in one mole of any element. We can use this conversion to find the total number of atoms in the sample.

$1.25 moles* \frac{6.022*10^{23}atoms}{1mol} = 7.5*10^{23} atoms$

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Example Question #1 : Using Avogadro's Number

How many atoms are there in 47.5g of boron?

Possible Answers:

$3.65*10^{22} \text{atoms}$

$2.65*10^{23} \text{atoms}$

$2.65*10^{24} \text{atoms}$

$5.15*10^{24} \text{atoms}$

Correct answer:

$2.65*10^{24} \text{atoms}$

Explanation:

To do this problem we have to first convert grams to moles, then moles to atoms using Avogadro's number:

$47.5g*\frac{1mol\ B}{10.8g\ B}*\frac{6.022*10^{23}\ \text{atoms}}{1mol}=2.65*10^{24}\text{atoms}$

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Example Question #1 : Using Avogadro's Number

How many moles of carbon are in a sample of $25.125*10^{27}$ atoms?

Possible Answers:

$4.17*10^{3} mol$

$4.17*10^{4} mol$

$1.51*10^{4} mol$

$1.51*10^{6} mol$

Correct answer:

$4.17*10^{4} mol$

Explanation:

To solve, we need to convert atoms to moles using Avogadro's number:

$25.125*10^{27}\text{atoms}*\frac{1mol}{6.022*10^{23}\text{atoms}}=4.17*10^4mol$

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Example Question #1 : Using Avogadro's Number

How many atoms of sodium are in three moles of $Na_{2}S$ ?

Possible Answers:

$3.61*10^{24} \text{atoms Na}$

$3.61*10^{26} \text{atoms Na}$

$3.61*10^{22} \text{atoms Na}$

$1.31*10^{24} \text{atoms Na}$

Correct answer:

$3.61*10^{24} \text{atoms Na}$

Explanation:

To answer this question, we have to find the number of moles of sodium in this compound. Note that there are two sodium atoms per molecule. Then, multiply the total moles of sodium by Avogadro's number.

$3 mol\ Na_{2}S*\frac{2mol\ Na}{1 mol\ Na_{2}S}*\frac{6.022*10^{23}\text{atoms\ Na}}{1 mole\ Na}=3.61*10^{24}\text{atoms Na}$

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Example Question #1 : Using Avogadro's Number

You have a neutral balloon. If you were to add 21,000 electrons to it, what would its net charge be?

$e=1.6\cdot 10^{-19}$ = charge of one electron

Possible Answers:

$-3.08\cdot 10^{-9}\mu C$

None of the other answers is correct

$3.08\cdot 10^{-9}\mu C$

$-3.36\cdot 10^{-9}\mu C$

$3.36\cdot 10^{-9}\mu C$

Correct answer:

$-3.36\cdot 10^{-9}\mu C$

Explanation:

The elemental charge is the magnitude of charge, in Coulombs, that each electron or proton has. Because electrons have a negative charge, don't forget to add a negative sign into the equation.

$q=-(\#\ of\ electrons)(e)$

$q= -(21000)(1.6\cdot 10^{-19})$

$q= -3.36\cdot 10^{-15}$

When you convert the answer to microcoulombs, the answer is $-3.36\cdot 10^{-9}\mu C$ :

$-3.36\cdot 10^{-15} C\cdot\frac{1.0\cdot10^6 \mu C}{1 C}= -3.36\cdot 10^{-9}\mu C$

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Example Question #1 : Using Avogadro's Number

What is the mass of $5.34\times10^{15}$ particles of $C_2H_6$ ?

Possible Answers:

$1.78\times 10^{-15}\hspace{1 mm}g\hspace{1 mm}C_2H_6$

$6.02\times 10^{-7}g\hspace{1 mm}C_2H_6$

$1.78\times 10^{-7}\hspace{1 mm}g\hspace{1 mm}C_2H_6$

$2.66\times 10^{-7}g\hspace{1 mm}C_2H_6$

$2.66\times 10^{-15}\hspace{1 mm}g\hspace{1 mm}C_2H_6$

Correct answer:

$2.66\times 10^{-7}g\hspace{1 mm}C_2H_6$

Explanation:

$5.34\times10^{15}\hspace{1 mm}particles\hspace{1 mm}C_2H_6\times \frac{1\hspace{1 mm}mole\hspace{1 mm}C_2H_6}{6.022\times 10^{23}\hspace{1 mm}particles\hspace{1 mm} C_2H_6}\times \frac{30.0\hspace{1 mm}g\hspace{1 mm}C_2H_60=}{1\hspace{1 mm}mole\hspace{1 mm}C_2H_6}=2.66\times 10^{-7}g\hspace{1 mm}C_2H_6$

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Example Question #1 : Moles And Molar Mass

How many atoms are in 1 mole of H₂?

Possible Answers:

1.2044 * 10²⁴

None of the other answers

3.626 * 10⁴⁷

6.022 * 10²³

6.022 * 10⁴⁶

Correct answer:

1.2044 * 10²⁴

Explanation:

This question requires an understanding of what avogadro's number actually represents. Avogadro's number, 6.022 * 10²³ is the number of things in one mole. The question indicates that there is 1 mole of H₂. Thus there are 6.022 * 10²³ molecules of H₂. However the question is asking for the amount of atoms in 1 mole of H₂. Thus we must consider the makeup of an H₂ molecule, where we see that it is a diatomic molecule. Thus we must multiply 6.022 * 10²³by 2 to calculate the number of individual atoms present in 1 mole of H₂. We find our answer to be 1.2044 * 10²⁴.

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Example Question #2 : Moles And Molar Mass

A chemist has $4.37*10^{-3}g$ of $CH_3CH_2OH$ .

How many molecules of $CH_3CH_2OH$ does she have?

Possible Answers:

$9.50*10^{19}$

$4.73*10^{19}$

$5.72*10^{18}$

$5.72*10^{19}$

$5.73*10^{19}$

Correct answer:

$5.72*10^{19}$

Explanation:

$4.37\times 10^{-3}g\hspace{1 mm}CH_3CH_2OH\times \frac{1\hspace{1 mm}mole\hspace{1 mm}CH_3CH_2OH}{46\hspace{1 mm}g\hspace{1 mm}CH_3CH_2OH}\times\frac{6.022\times 10^{23}\hspace{1 mm}molecules\hspace{1 mm} CH_3CH_2OH}{1\hspace{1 mm}mole\hspace{1 mm}CH_3CH_2OH}=5.72\times10^{19}\hspace{1 mm}molecules\hspace{1 mm}CH_3CH_2OH$

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Example Question #1 : Using Avogadro's Number

How many hydrogen atoms are in 46.3 g of ethanol, CH_6O ?

Possible Answers:

$6.0*10^{23}\ \text{atoms}$

$3.1*10^{24}\ \text{atoms}$

$4.9*10^{24}\ \text{atoms}$

$2.7*10^{25}\ \text{atoms}$

Correct answer:

$4.9*10^{24}\ \text{atoms}$

Explanation:

To determine the number of hydrogen atoms, divide the mass of ethanol by its molar mass to get moles of ethanol.

$MM=(12)+6(1)+(16)=34\frac{g}{mol}$

$46.3g*\frac{1mol}{34g}=1.36mol\ CH_6O$

Multiply this by six atoms of hydrogen per molecule of ethanol and by Avogadro's number to get the number of hydrogen atoms.

$1.36mol\ CH_6O*\frac{6mol\ H}{1mol\ CH_6O}=8.17mol\ H$

$8.17mol\ H*\frac{6.022*10^{23}\text{atoms}}{1mol}=4.9*10^{24}\text{atoms}$

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Example Question #4 : Moles And Molar Mass

How many hydrogen atoms are present in 500 mL of water at room temperature?

$\rho_{\text{water}}=1\frac{g}{mL}$

Possible Answers:

$3.35*10^{25}\ \text{atoms}$

$5.96*10^{26}\ \text{atoms}$

$1.67*10^{25}\ \text{atoms}$

$5.01*10^{25}\ \text{atoms}$

Correct answer:

$3.35*10^{25}\ \text{atoms}$

Explanation:

Use the density of water, the molar mass of water, and Avogadro's number to calculate the number of molecules of water.

We have 500mL of water. Use the density to convert this to grams; then use the molar mass of water to convert this to moles.

$500mL*\frac{1g}{1mL}*\frac{1mol}{18g}=27.78mol\ H_2O$

There are two moles of hydrogen atoms per one mole of water.

$27.78mol\ H_2O*\frac{2mol\ H}{1mol\ H_2O}=55.56mol\ H$

Finally, multiply by Avogadro's number.

$55.56mol\ H*\frac{6.022*10^{23}\text{atoms}}{1mol}=3.35*10^{25}\text{atoms}$

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High School Chemistry : Using Avogadro's Number

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Using Avogadro's Number

Example Question #1 : Using Avogadro's Number

Example Question #1 : Using Avogadro's Number

Example Question #1 : Using Avogadro's Number

Example Question #1 : Using Avogadro's Number

Example Question #1 : Using Avogadro's Number

Example Question #1 : Moles And Molar Mass

Example Question #2 : Moles And Molar Mass

Example Question #1 : Using Avogadro's Number

Example Question #4 : Moles And Molar Mass

All High School Chemistry Resources

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