Mole Problems

SaiGuruCool

This is a collection of several problems regarding moles, empirical formulas, percent composition, conversions from moles to grams, mole to volume, grams to volume, etc. All work is shown with brief explanations.

COLLECTION OF MOLE PROBLEMS

How many atoms are present in 9.425 grams of Boron?

Molar mass of Boron from Periodic Table = 10.8 g/mol (rounded to 1 decimal place)

One mole of boron = 6.02 x 10^23 atoms

(9.425 grams) * [(1 mole of boron)/(10.8 grams)] = 0.8726851852… (do not round intermediately)

0.8726851852… moles * [(6.02 x 10^23 atoms)/(1 mole)] = 5.2536 x 10^23 moles (round to 4 sig figs since data given in problem has 4 sig figs)

How many moles of tetraphosphorus decoxide does a 250.0 g sample represent?

Molar mass of phosphorus (P) = 31.0 grams (rounded to 1 decimal place)

Molar mass of oxygen (O) = 16.0 grams (rounded to 1 decimal place)

Molar mass of tetraphosphorus decoxide from the periodic table = (4 * 31.0 grams) + (10 * 16.0 grams) = 284.0 g/mol

250 g * [(1 mole)/(284 g/mol)] = 0.8803 moles (round to 4 sig figs since data given in problem has 4 sig figs)

How do you find the empirical formula of the substance that is 80.38% bismuth, 18.46% oxygen and 1.16% hydrogen?

First convert these percentages into moles

Molar mass of Bismuth = 209.0 g/mol

80.38% Bismuth - 80.38 grams * [(1 mole)/(209.0 g/mol)] = 0.3845933014... moles (do not round intermediately)

Molar mass of Oxygen = 16.0 g/mol

18.46% Oxygen - 18.46 grams * [(1 mole)/(16.0 g/mol)] = 1.15375 moles (do not round intermediately)

Molar mass of Hydrogen = 1.0 g/mol

1.16% Hydrogen - 1.16 grams * [(1 mole)/(1.0 g/mol)] = 1.16 moles (do not round intermediately)

Divide each moles by the smallest mole conversion:

COLLECTION OF MOLE PROBLEMS

Bismuth has the smallest amount of moles

Bismuth - (0.3845933014... moles)/(0.3845933014... moles) = 1 atoms

Oxygen - (1.15375 moles)/(0.3845933014... moles)  = 2.999922244 atoms

Hydrogen - (1.16 moles)/(0.3845933014... moles) = 3.016173177 atoms

Formula:

Bi1O2.999922244H3.016173177

Round to the whole numbers:

Bi1O3H3

A compound was analyzed and was found to contain the following percentages by mass: phosphorus, 90.10%; hydrogen 8.90%. Determine the empirical formula of the compound.

First convert these percentages into moles

Molar mass of Phosphorus = 31.0 g/mol

90.10% Phosphorus - 90.10 grams * [(1 mole)/(31.0 g/mol)] = 2.906451613... moles (do not round intermediately)

Molar mass of Hydrogen = 1.0 g/mol

8.90% Hydrogen - 8.90 grams * [(1 mole)/(1.0 g/mol)] = 8.90 moles (do not round intermediately)

Divide each moles by the smallest mole conversion:

Phosphorus has the smallest amount of moles

Phosphorus - (2.906451613... moles)/(2.906451613... moles) = 1 atoms

Hydrogen - (8.90 moles)/(2.906451613... moles) = 3.062153163 atoms

Formula:

P1H3.062153163

Round to the nearest whole number:

P1H3 – Phosphorus Tetrahydride

COLLECTION OF MOLE PROBLEMS

Analysis of an unknown compoud gave the following mass percentages:

17.5% Na,39.7%Cr,and 42.8% 0.

What is the empirical Formula?

First convert these percentages into moles

Molar mass of Sodium = 23.0 g/mol

17.5% Sodium - 17.5 grams * [(1 mole)/(23.0 g/mol)] = 0.7608695652... moles (do not round intermediately)

Molar mass of Chromium = 52.0 g/mol

39.7% Chromium - 39.7 grams * [(1 mole)/(52.0 g/mol)] = 0.5711538462... moles (do not round intermediately)

Molar mass of Oxygen = 16.0 g/mol

42.8% Oxygen - 42.8 grams * [(1 mole)/(16.0 g/mol)] = 2.675 moles (do not round intermediately)

Divide each moles by the smallest mole conversion:

Chromium has the smallest amount of moles

Sodium - (0.7608695652... moles)/(0.5711538462... moles) = 0.9966049721 atoms

Chromium - (0.5711538462... moles)/(0.5711538462... moles)  = 1 atom

Oxygen - (2.675 moles)/(0.5711538462... moles) = 3.503778338 atoms

Formula:

Na0.9966949721Cr1O3.503778338

Round to the nearest whole number. However, if one of the elements has about n and a half atoms, then multiply each number atoms by 2:

Na1Cr1O3.503778338 --> Na2Cr2O7.007556675 --> Na2Cr2O7