The mole fraction is:

moles of target substance divided by total moles involved

The symbol for the mole fraction is the lower-case Greek letter chi, χ. You will often see it with a subscript: χ_{solute} is an example.

**Example #1:** 0.100 mole of NaCl is dissolved into 100.0 grams of pure H_{2}O. What is the mole fraction of NaCl?

**Solution:**

100.0 g / 18.0 g mol¯^{1}= 5.56 mol of H_{2}OAdd that to the 0.100 mol of NaCl = 5.56 + 0.100 = 5.66 mol total

Mole fraction of NaCl = 0.100 mol / 5.66 mol = 0.018

What is the mole fraction of the H_{2}O?

5.56 mol / 5.66 mol = 0.982

By the way, another way to figure out the last substance is 1.00 minus (the total of all other mole fractions). In this case 1.00 - 0.018 = 0.982. Remember that all the mole fractons in the solution should total up to one.

Notice that the mole fraction has no units on it and is written as a decimal value. Do not change it to percent.

Note of caution: you could see the term "mole percent." It is simply the mole fraction mltiplied by 100. For example, in the problem just below, the mole fraction of cinnamic acid is 0.2885. Its mole percent would be 28.85%.

The ChemTeam advises against the use of the tem "mole percent." However, do what your teacher desires you to do.

**Example #2:** A solution is prepared by mixing 25.0 g of water, H_{2}O, and 25.0 g of ethanol, C_{2}H_{5}OH. Determine the mole fractions of each substance.

**Solution:**

1) Determine the moles of each substance:

H_{2}O ⇒ 25.0 g / 18.0 g/mol = 1.34 mol

C_{2}H_{5}OH ⇒ 25.0 g / 46.07 g/mol = 0.543 mol

2) Determine mole fractions:

H_{2}O ⇒ 1.34 mol / (1.34 mol + 0.543 mol) = 0.71

C_{2}H_{5}OH ⇒ 0.543 mol / (1.34 mol + 0.543 mol) = 0.29

**Example #3:** A solution contains 10.0 g pentane, 10.0 g hexane and 10.0 g benzene. What is the mole fraction of hexane?

**Solution:**

1) You need to determine the moles of pentane, hexane and benzene:

to do this, you need the molecular weights. Here are the formulas:pentane: C_{5}H_{12}

hexane: C_{6}H_{14}

benzene: C_{6}H_{6}

2) When you have the moles of each, add them together.

3) Then, divide the moles of hexane by the total.

**Example #4:** The molality of an aqueous solution of sugar (C_{12}H_{22}O_{11}) is 1.62*m*. Calculate the mole fractions of sugar and water.

**Solution:**

1) Molality is moles solute / kg of solvent. Therefore we know our solution is:

1.62 mol C_{12}H_{22}O_{11}

1.00 kg = 1000 g of water

2) Calculate the moles of water present:

1000 g / 18.0152 g/mol = 55.50868 mol

3) Determine the mole fraction of the sugar:

1.62 mol / (1.62 mol + 55.50868 mol) = 0.028357 = 0.0284 (to three sf)

4) You can calculate the mole fraction of the water by subtraction.

**Example #5:** How many grams of water must be used to dissolve 100.0 grams of sucrose (C_{12}H_{22}O_{11}) to prepare a 0.020 mole fraction of sucrose in the solution?

**Solution:**

1) Determine moles of sucrose:

100.0 g / 342.2948 g/mol = 0.292145835 mol

2) Determine moles of water required to make the solution 0.020 mole fraction of sucrose:

0.020 = 0.292 / (0.292 + x)(0.020) (0.292 + x) = 0.292

0.00584 + 0.02x = 0.292

0.02x = 0.28616

x = 14.308 mol of H

_{2}OComment: you can also do this:

0.292 is to 0.02 as x is to 0.98

3) Determine grams of water:

14.308 mol x 18.015 g/mol = 258.0 g

**Example #6:** Surprisingly, water (in the form of ice) is slightly soluble in liquid nitrogen. At -196 °C, (the boiling point of liquid nitrogen) the mole fraction of water in a saturated solution is 1.00 x 10^{-5}. Compute the mass of water that can dissolve in 1.00 kg of boiling liquid nitrogen.

**Solution:**

1) Use the definition of mole fraction to set up the following:

χ_{water}= moles water / (moles water + moles nitrogen)1.00 x 10

^{-5}= x / (x + 71.3944041)I'm going to carry some guard digits until the end of the calculation.

2) Some algebra:

(1.00 x 10^{-5}) (x) + 7.139440411 x 10^{-4}= x0.99999x = 7.139440411 x 10

^{-4}x = 7.139511806 x 10

^{-4}mol of H_{2}O

3) Calculate grams of water from moles of water:

7.139511806 x 10^{-4}mol x 18.0152 g/mol = 1.2862 x 10^{-2}g1.29 x 10

^{-2}g (to three sf)

**Example #7:** What is the mole fraction of cinnamic acid in a mixture that is 50.0% weight urea in cinnamic acid (urea = 60.06 g/mol; cinnamic acid = 148.16 g/mol)

**Solution:**

1) Let us assume 100.0 g of this mixture are present. Therefore:

50.0 g is urea

50.0 g is cinnamic

2) Convert grams to moles:

urea: 50.0 g / 60.06 g/mol = 0.8325 mol

cinnamic acid: 50.0 g / 148.16 g/mol = 0.3375 mol

3) Determine mole fraction of cinnamic acid:

0.3375 mol / 1.1700 mol = 0.2885

**Bonus Example:** You are given the following information about a mixture of KCl, NaCl and LiCl:

(a) the mass of KCl + NaCl is 2.370 g

(b) the mass of NaCl + LiCl is 1.290 g

(c) the mass of AgCl precipitated from the mixture is 6.435 g

Determine the mole fraction of LiCl in the mixture.

**Solution:**

1) The mass of KCl in the mixture will be our unknown, 'x.' This leads a description of the mass of NaCl present and the mass of LiCl present:

mass of NaCl in the mixture ---> 2.370 - x

mass of LiCl in the mixture ---> 1.290 - (2.370 - x) ---> x - 1.080

2) Change all masses to moles:

KCl ---> x / 75.00

NaCl ---> (2.370 - x) / 58.00

LiCl ---> (x - 1.080) / 42.00AgCl ---> 6.432 g / 143.0 g/mol = 0.044979 mol

3) We know the following to be true:

moles KCl + moles NaCl + moles LiCl react to produce moles AgCl[x / 75.00] + [(2.370 - x) / 58.00] + [(x - 1.080) / 42.00] = 0.044979 mol

Algebra!

x = 1.500 g

4) We can now determine the mass of each substance in the mixture:

KCl ---> 1.500 g

NaCl ---> 2.370 - 1.500 = 0.870 g

LiCl ---> 1.500 - 1.080 = 0.420 g

5) Determine the moles of each substance:

KCl ---> 1.500 g / 75.00 g/mol = 0.0200 mol

NaCl ---> 0.870 g / 58.00 g/mol = 0.0150 mol

LiCl ---> 0.420 g / 42.00 g/mol = 0.0100 mol

6) Determine the mole fraction of LiCl:

0.0200 mol + 0.0150 mol + 0.0100 mol = 0.0450 mol0.0100 mol / 0.0450 mol = 0.222 (to three sig figs)