The atomic mass of an element is determined by the proportional mass of each elemental isotope. We know that there are only two isotopes of lithium; therefore, their percentages must add to 100%.

$\displaystyle ^6Li=x\ \text{and}\ ^7Li=y$

$\displaystyle x+y=1$

The atomic mass will be equal to the mass of each isotope multiplied by its abundance.

$\displaystyle 6.941amu=x(6)+y(7)$

We can substitute an algebraic expression to solve for one of our variables.

$\displaystyle x+y=1\rightarrow y=1-x$

$\displaystyle 6.941amu=x(6)+(1-x)(7)$

$\displaystyle 6.941amu=6x+7-7x$

$\displaystyle x=0.059$

Using this value, we can solve for the abundance of the other isotope.

$\displaystyle y=1-x\rightarrow y=1-(0.059)=0.941$

Converting these values to percentages gives us our final answer.

$\displaystyle ^7Li:94.1\%\ ,\ ^6Li:5.9\%$

It is important to remember that one mole of an element contains 6.022 * 10²³ atoms. Since there are two moles of carbon dioxide (CO₂), we can conclude that there are two moles of carbon. As a result, there are $\displaystyle \small 2mol(6.022*10^{23}\frac{atoms}{mol}) =$$\displaystyle \small 1.2*10^{24}$ atoms.

A mole is defined as the number of atoms present in $\displaystyle 12g$ of carbon-12. Obtaining a $\displaystyle 12g$sample of carbon-12 will give you $\displaystyle 1\: mole$ of carbon-12. You can also find the number of moles by multiplying the mass by the molecular weight of carbon:

$\displaystyle (12g)(\frac{1mol}{12.001g}) = 1\: mol\: C$

Remember that a mole of any element contains $\displaystyle 6.022*10^{23}$ atoms (Avogadro’s number). Since we have exactly one mole in the sample, there will be exactly $\displaystyle 6.022*10^{23}$ atoms of carbon-12.

Remember that the empirical formula relies on the ratio of the moles of elements. To get the number of atoms, you would have to multiply the moles of each element by the Avogadro’s number ($\displaystyle 6.022*10^{23}$). You would use this number for every element (Avogadro’s number doesn’t change for each element). This means that the ratio of the number of atoms will be the same as the ratio of the number of moles, and you will get the same empirical formula. There is a constant relationship between the number of moles and the number of atoms in a sample.

Each answer choice is presented in grams. To convert to moles, we will need to divide each choice by the atomic mass of the presented element.

$\displaystyle 2g\ H*\frac{1mol}{1g\ H}=2mol\ H$

$\displaystyle 5g\ Cl*\frac{1mol}{35.5g\ Cl}=0.14mol\ Cl$

$\displaystyle 10g\ Ag*\frac{1mol}{108g\ Ag}=0.09mol\ Ag$

$\displaystyle 20g\ Au*\frac{1mol}{197g\ Au}=0.10mol\ Au$

$\displaystyle 3g\ Li*\frac{1mol}{7g\ Li}=0.43mol\ Li$

Two grams of hydrogen atoms will result in the greatest amount of moles.

Avogadro's number is used to convert between moles and atoms (or molecules).

$\displaystyle \frac{6.022*10^{23}\ \text{molecules/atoms}}{1\ \text{mole}}$

This immediately eliminates two answer choices, since moles and molecules in a sample are both contained in the constant. Atoms in a sample of diatomic gas is also easily related to Avogadro's number, as each molecule in the sample will contain exactly two atoms. Avogadro's number can be used to determine the number of moles, molecules, or atoms in a sample of diatomic gas.

Converting grams in a sample to moles allows us to use Avogadro's number to further convert to molecules.

Partial pressure of a gas is directly related to mole fraction, but cannot be used to determine the moles of gas unless the total moles in the sample is known. Partial pressure represents a relationship between the sample and its particular environment, whereas constants govern conversion between moles, grams, atoms, and molecules. Since partial pressure is not directly related to these terms by a constant, Avogadro's number cannot be applied to partial pressure to determine any useful data.

To solve, we will need to find the mass of one liter of DEET and use the mass percentage of nitrogen to find the mass of nitrogen represented. Then, the atomic mass of nitrogen can be used to convert this to moles, and Avogadro's number can be used to convert to atoms.

First, find the mass of DEET in a one-liter sample:

$\displaystyle 1L*\frac{1000cm^3}{1L}*\frac{1m^3}{(100)^3cm^3}*\frac{998kg}{1m^3}=0.998kg$

Find the mass of nitrogen in 998 grams of DEET by finding the percentage of nitrogen by mass:

$\displaystyle \%\ N=\frac{mass\ N}{total\ mass}*100\%$

$\displaystyle C_{12}H_{17}ON$

$\displaystyle \frac{14 \frac{g}{mol}}{191 \frac{g}{mol}} * 100 = 7.3% N$

Use this mass percentage and atomic mass to find the moles of nitrogen in the one-liter sample.

$\displaystyle 998g\ DEET*\frac{7.3g\ N}{100g\ DEET}*\frac{1mol\ N}{14.0g\ N}=5.20mol\ N$

Use Avogadro's number to find the number of atoms in this sample.

$\displaystyle 5.20mol*\frac{6.022*10^{23}atoms}{1mol}=3.1*10^{24}atoms$

An alternative method would be to convert the grams of DEET to moles, and then from moles to molecules. There is one nitrogen atom per molecule of DEET, so this would method also get you the same answer.

The number of atoms in a sample is dependent on the molar quantity, NOT the mass of the sample. Avogadro's number shows that there are $\displaystyle 6.022*10^{23}$ atoms in a mole of any element. Knowing this, we can find out how many moles of the element are represented by the given amount of atoms.

$\displaystyle \frac{1.31*10^{23}\text{atoms}}{6.022*10^{23}\frac{\text{atoms}}{mol}}=0.217mol$

By dividing the mass of the samples by the molar mass of each respective element, we can find which sample has this molar quantity.

Sodium: $\displaystyle 5g*\frac{1mol}{23g}=0.217mol$

Lithium: $\displaystyle 5g*\frac{1mol}{7g}=0.714mol$

Potassium: $\displaystyle 5g*\frac{1mol}{39g}=0.128mol$

As a result, a five-gram sample of sodium has $\displaystyle 1.31*10^{23} atoms$.

A full liter of a one molar solution of sodium hydroxide would contain one mole of sodium ions, or $\displaystyle \small 6.022*10^{23}$ ions. Here, you have only one tenth the volume, so multiply the number in one mole by one tenth.

$\displaystyle \small 10^{-1}*(6.022*10^{23 })= 6.022*10^{22}ions$

Now that we have reduced the volume, we need to account for the concentration.

$\displaystyle 0.023mM=0.023*10^{-3}M$

$\displaystyle (6.022*10^{22}ions)*(0.023*10^{-3}M)=1.385*10^{18}ions$

When finding the empirical formula for a compound, it helps to imagine a 100g sample of the molecule. This way, the percentages of the atoms can be converted to amounts in grams. At this point, dividing each amount by the atom's molar mass results in the values that will be compared to one another in order to find the ratio of atoms in the molecule.

$\displaystyle \small \frac{41.4}{12} = 3.45$ (carbon)

$\displaystyle \small \frac{6.9}{1} = 6.9$ (hydrogen)

$\displaystyle \small \small \frac{27.6}{16} = 1.72$ (oxygen)

$\displaystyle \small \frac{24.1}{14} = 1.72$ (nitrogen)

From these calculations, we can set up a ratio.

3.45C : 6.9H : 1.72O : 1.72N

If we divide this ratio by the smallest value (1.72), we can see that it can be reduced.

2C : 4H : 1O : 1N

As a result, the empirical formula for the compound is C₂H₄ON.