AP Chemistry - Moles and Molar Mass

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

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.

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

What is the molar mass of ?

$164.1\frac{g}{mol}$

$102.09\frac{g}{mol}$

$204.18\frac{g}{mol}$

$116.1\frac{g}{mol}$

$150.09\frac{g}{mol}$

Explanation

The chemical formula has a 2 right outside of the parentheses surrounding nitrate, so we need to distribute/multiply that 2 to everything inside the parentheses. That means that we have 1 calcium, two nitrogens, and six oxygens, and we need to add up all those masses to get the molar weight.

$MW=40.08g+(2\cdot 14.01g)+ (6\cdot 16g)$

$MW=164.1\frac{g}{mol}$

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

How many molecules of $CH_3CH_2OH$ does she have?

$5.72*10^{19}$

$9.50*10^{19}$

$5.72*10^{18}$

$5.73*10^{19}$

$4.73*10^{19}$

Explanation

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

What is the molar mass of ammonium hydroxide?

$35.05\frac{g}{mol}$

$32.48\frac{g}{mol}$

$40.09\frac{g}{mol}$

$18.042\frac{g}{mol}$

$30.01\frac{g}{mol}$

Explanation

The formula for ammonium hydroxide is . To find molar mass, add up the individual masses of each element, and remember to multiply that mass by the molar ratio of that element.

$14.01+(5\cdot 1.008)+16 = 35.05\frac{g}{mol}$

A ring of imitation jewelry is coated with $1.00: \mu g$ of silver which is equivalent to:

of or $5.58\times 10^{15}$ atoms of

$6.02\times 10^{23}$ atoms of

atoms of

$5.58\times 10^{-15}$ atoms of

Explanation

To solve this problem we use the concept of atomic mass and Avogadro's number:

$1.00 : \mu g:; of: Ag\times \frac{10^{-6}g: of: Ag}{1: \mu g: of: Ag}\times\frac{1: mol: Ag}{107.87: g: of: Ag}\times$

$\times \frac{10^{9}: nmol : of: Ag}{1: mol: Ag}=9.27: nmol: of: Ag$

And the number of Ag atoms will be:

$1.00 : \mu g:; of: Ag\times \frac{10^{-6}g: of: Ag}{1: \mu g: of: Ag}\times\frac{1: mol: Ag}{107.87: g: of: Ag}\times$

$\times \frac{6.02\times 10^{23}: atoms: of: Ag}{1: mol: of: Ag}=5.58\times 10^{15}: atoms : of: Ag$

What is the molar mass of oxalic acid?

$30.068\frac{g}{mol}$

$27.99\frac{g}{mol}$

$18.058\frac{g}{mol}$

$29.06\frac{g}{mol}$

$17.05\frac{g}{mol}$

Explanation

First, write the formula for oxalic acid: . We know that this formula is correct because when we have an acid with an "-ic" ending and no "hydro-" prefix, that means it's coming from a polyatomic ion with an "-ate" ending. So we know that oxal"ic" acid is derived from the oxal"ate" polyatomic ion. The formula for that ion is $C_2O_4^{2-}$ . Putting a hydrogen atom in front of this ion, and balancing the charges, we get .Then add the atomic masses of each of the individual elements in their respective ratios, to get the molar mass for the whole compound.

$(2\cdot 1.008)+(2\cdot 12.01)+(2\cdot 16)= 30.068\frac{g}{mol}$

How many molecules are there in of $BaSO_{4}$ ?

$3.3 * 10^{24} molecules\ BaSO_{4}$

$3.3121 * 10^{24} molecules\ BaSO_{4}$

$3.312 * 10^{24} molecules\ BaSO_{4}$

$3.31 * 10^{24} molecules \ BaSO_{4}$

$3.13*10^{23}molecules \ BaSO_4$

Explanation

Use dimensional analysis to convert from moles to molecules.

$5.5 mol \ BaSO_{4} * \frac{6.022*10^{23} \ molecules}{1 mol\ BaSO_4}= 3.3*10^{24} molecules\ BaSO_4$

Since the given measurement of has two significant figures, the answer must also have two significant figures.

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

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

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

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

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

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.

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

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

What is the percent by mass of the anion in lithium iodate, $LiIO_{4}$ ?

This compound isn't ionic.

Explanation

Lithium iodate is comprised of $Li^{+1}$ and $(IO_{4})^{-1}$ . The anion refers to the negative ion, so we are looking for the percent by mass of $(IO_{4})^{-1}$ .