Moles and Molar Mass - AP Chemistry

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Question

The empirical formula of an unknown compound is CH2. Which of the following could not be the molar mass of the unknown compound?

Answer

An empirical formula is a way of expressing a chemical formula in the smallest possible ratio. The compound C3H6 has an empirical formula of CH2 because the simplest ratio between this compound's constituent elements is one carbon for every two hydrogens. This empirical formula can represent a number of compounds, but we know that each compound has a molar mass that is a multiple of the molar mass of the empirical formula. The empirical formula has a molar mass of 14 g/mol so the molar masses of compounds that can be represented by this empirical formula are (28,42,56 . . .).

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Question

How many atoms are in 1 mole of H2?

Answer

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 * 1023 by 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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Question

What is the molar mass of ?

Answer

To find the molecular weight, you need to add together the atomic masses of each of the atoms. 1 sulfur is 32g/mol, 1 oxygen is 16g/mol, and 2 chlorine is 71g/mol (each Cl atom is 35.5g/mol). These values are taken from the periodic table. Added together, they make 119g/mol.

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Question

Which of the following compounds has a formula weight within 74–75 g/mol?

Answer

PCl5, BF3, and C6H12O6 are not ionic compounds, and can thus be eliminated because "formula weight" implies that we are dealing with an ionic compound. This leaves only the correct answer, KCl, which can also be verified as correct by calculating the weight. K is 39g, while Cl is 35.5g. Together, this gives a total weight of approximately 74.5g.

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Question

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

How many molecules of $CH_3CH_2OH$ does she have?

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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.

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

$8.17mol\ H\frac{6.02210^{23}\text{atoms}}{1mol}=4.9*10^{24}\text{atoms}$

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Question

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

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

Answer

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.

$55.56mol\ H\frac{6.02210^{23}\text{atoms}}{1mol}=3.3510^{25}\text{atoms}$

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Question

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

Answer

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 $(IO_{4})^{-1}$ .

The total molar mass of lithium iodate is $197.84\frac{g}{mol}$ .

$(6.94\frac{g}{mol}Li)+(126.90\frac{g}{mol}I)+4(16.0\frac{g}{mol}O)=197.84\frac{g}{mol}$

The anion, $(IO_{4})^{-1}$ , contributes 190.9g to the total mass.

$(126.9\frac{g}{mol}I)+4(16\frac{g}{mol}O)=190.9\frac{g}{mol} IO_{4}^{-1}$

Find the percent by mass by dividing the mass of the anion by the total molecular mass.

$\frac{190.9g\ IO_{4}}{197.841g\ total}=96.49%$

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Question

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

Answer

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:

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

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

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Question

What is the molar mass of ?

Answer

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.

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

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

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Question

Avogadro's number can convert all of the following units into moles, and vice versa, except __________.

Answer

Avogadro's number can change: atoms, molecules, particles, formula units, and photons into moles (and vice versa). But in order to change grams to moles or moles to grams, the molar mass is needed.

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Question

What is the molar mass of potassium sulfide?

Answer

In order to find the molar mass of potassium sulfide, you must first write out the chemical formula for potassium sulfide. The symbol for the potassium ion is , and the symbol for sulfur is . We know that this is an ionic compound because we have a metal involved (potassium) with a nonmetal (sulfur), so we must balance the net charge while making sure both atoms have a full octet. Since potassium is in group I, we know that it can easily lose one electron to fill its outer shell of electrons. Sulfur is in group VI, so we know that it can easily gain two electrons to fill its valence shell of electrons. It would take two potassium atoms to donate one electron each to sulfur to make this compound stable. Thus the chemical formula would be . Use the periodic table to find the molar mass.

$MM=(2\cdot 39.10g) +32.07g= 110.27\frac{g}{mol}$

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Question

Nuclear power has the potential to supply large amounts of energy from a relatively small amount of starting material (when compared to combustion of fossil fuels). Nuclear fuels have different yields of energy per unit mass (as well as radioactive wastes). For the following problem assume the following:

1 kilowatt hour = $2.25*10^{19} MeV$

United states energy consumption per capita per year = 13246.00kWH

The energy released in the nuclear fission of one atom of uranium-235 is 211.30MeV.

The natural abundance of uranium-235 is 0.72%

The energy released from the fission of one atom of thorium-232 is 191.00MeV (Involves several steps known as the thorium fuel cycle).

Thorium-232 is the most abundant isotope of thorium and for this problem can be considered to be 100%

Assume that 100% of the U-235 and Th-232 can be converted into energy.

How many grams of thorium or uranium (of all isotopes mined from natural sources) must be mined to provide the energy consumption needs of one average American in a year (given the above assumptions)?

Note: solve this problem twice; assuming 100% of energy consumption is produced from thorium in the first case and uranium in the second.

Answer

To solve this problem we need to do some stoichiometry. Once we have gathered all the conversion information needed it becomes a matter of stringing the equation together and making sure our units cancel out to give us the units desired. Trying to tackle this problem stepwise could soon become confusing as there are more than a few steps involved.

Calculation for the Mass of Uranium Mined to Supply the Energy Needs of 1 American for a Year:

$\frac{13246.00kWh}{\frac{1\ American}{year}}\frac{2.210^{19}MeV}{1kWh}* \frac{1atom{{}^{235}\textrm{U}}}{211.30MeV}\frac{100atomsU}{0.72atoms{{}^{235}\textrm{U}}}\frac{1molU}{6.02210^{23}atomsU}$

$ \frac{238.03gU}{1molU}=\frac{75.71gU}{1\ American\ year^{-1}}$

It is worth noting that this is the mass of U and doesn't account for the fact that the U mined is not present as a pure element and therefore in reality the mass of material mined overall to get that 75.71g is much greater.

Calculation for the mass of Thorium Mined to Supply the Energy Needs of 1 American for a Year:

$\frac{13246.00kWh}{\frac{1\ American}{year}}\frac{2.210^{19}MeV}{1kWh}* \frac{1atom{{}^{232}\textrm{Th}}}{191.00MeV}\frac{1atomTh}{1atom{{}^{232\textrm{Th}}}}\frac{1molTh}{6.022* 10^{23}atomsTh}}$

$*\frac{232.04gTh}{1molTh}=\frac{0.59gTh}{1American/ yr^{-1}}$

The mass of Thorium mined is much less than the mass of Uranium mined mainly because of the concept of isotopic abundance. The isotope of thorium that is useful as nuclear fuel accounts for practically 100% of the naturally occurring thorium, whereas the useful isotope of uranium accounts for only 0.72% of the naturally occurring uranium. Also it is worth noting that like uranium, naturally occurring thorium is not present as a pure compound so therefore in reality the mass of material mined to get that 0.59g is also much greater. However it is also worth noting that thorium is much more abundant than uranium.

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Question

What is the molar mass of ammonium hydroxide?

Answer

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.

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

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Question

What is the molar mass of oxalic acid?

Answer

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.

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

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Question

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

Answer

Use dimensional analysis to convert from moles to molecules.

$5.5 mol \ BaSO_{4} * \frac{6.02210^{23} \ molecules}{1 mol\ BaSO_4}= 3.310^{24} molecules\ BaSO_4$

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

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Question

How many moles are in 610 atoms of silver?

Answer

$610 atoms Ag * \frac{1 mol Ag}{6.022 * 10^{23} atoms Ag}= 1.0 * 10^{-21} mols Ag$

The reason the answer has 2 significant figures, is because the given measurement has 2 significant figures. Remember: in non-decimal numbers, zeros only count if in between non-zero numbers.

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Question

A pure sample of $KClO_{3}$ is found to contain 71g of chlorine atoms. What is the mass of the sample?

Answer

First, find how many moles of chlorine are present in the sample:

$moles\ Cl = \frac{71g}{35.5\frac{g}{mol}} = 2 mol$

Because the ratio of moles of to moles of $KClO_{3}$ is 1:1, there are also 2 moles of $KClO_{3}$ . The next step is to find the mass of 2 moles of $KClO_{3}$ .

$2mol\ KClO_3 *{122.55\frac{g}{mol}}=245g$

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Question

A chemist has found that 2.0 moles of a mysterious compound contained 48g of carbon, 8g of hydrogen, and 64g of oxygen. What is the molecular formula of the compound?

Answer

Find the number of moles of each element.

$moles = \frac{grams}{MW}$

$\frac{48g\ C}{\frac{12g}{1mol}} = 4 mol \ C$

$\frac{8g \ H}{\frac{1g}{1mol}} = 8 mol \ H$

$\frac{64g\ O}{\frac{16g}{1mol}} = 4 mol\ O$

If the chemist has 2.0 moles of the compound, 1 mole would contain half as much. So 1mol of carbon, 4mol of hydrogen, and 2mol of oxygen, which forms acetic acid $(CH_{3}COOH)$

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