How many moles is in 9 g of H2O?

Water, which has the chemical formula H2O, is one of the most common chemical compounds on Earth. It is a molecule made up of 2 hydrogen atoms bonded to 1 oxygen atom. The mass of 1 mole of any substance is equal to its molecular weight in grams. So to figure out how many moles are in a given mass of a substance, you just need to know its molecular weight.

Table of Contents

Calculating the Molecular Weight of H2O

The molecular weight of a compound is equal to the sum of the atomic weights of each element in the compound. Looking at the periodic table:

Hydrogen (H) has an atomic weight of 1.008 g/mol

Oxygen (O) has an atomic weight of 16.00 g/mol

So for H2O:

2 hydrogen atoms with atomic weight 1.008 g/mol each = 2.016 g/mol

1 oxygen atom with atomic weight 16.00 g/mol

Adding these together:

Molecular weight of H2O = 2.016 g/mol + 16.00 g/mol = 18.02 g/mol

Converting Grams to Moles

Now that we know the molecular weight of H2O is 18.02 g/mol, we can use this to convert from grams of H2O to moles of H2O. The formula is:

Moles = Mass (g) / Molecular Weight (g/mol)

We are given that we have 9 g of H2O. Using the formula:

Moles of H2O = 9 g / 18.02 g/mol = 0.5 moles

Using a Dimensional Analysis Approach

Another way to do this conversion is using dimensional analysis, which allows you to cancel units:

9 g H2O x (1 mol H2O / 18.02 g H2O) = 0.5 mol H2O

This gives the same result of 0.5 moles of H2O in 9 grams of H2O.

Visualizing the Amount of H2O Molecules

It can help to visualize how many molecules are actually in 0.5 moles of H2O. Using Avogadro’s number which is 6.022×10^23 molecules per mole, we can calculate:

0.5 mol H2O x (6.022×10^23 molecules H2O/1 mol H2O) = 3.011×10^23 molecules of H2O

That’s over 30 sextillion water molecules in just 9 grams! Seeing the massive number of molecules helps grasp just how many particles are contained in a mole.

How Many Atoms Are in 0.5 Moles of H2O?

We can take this one step further to figure out how many total hydrogen and oxygen atoms are present in 0.5 moles of H2O.

Since one mole of H2O contains 2 moles of hydrogen and 1 mole of oxygen, in 0.5 moles of H2O there will be:

2 * 0.5 = 1 mole of hydrogen
1 * 0.5 = 0.5 moles of oxygen

Using Avogadro’s number again:

1 mole H x (6.022×10^23 atoms H/1 mole H) = 6.022×10^23 hydrogen atoms
0.5 moles O x (6.022×10^23 atoms O/1 mole O) = 3.011×10^23 oxygen atoms

So in summary, 0.5 moles of H2O contains:

6.022×10^23 hydrogen atoms
3.011×10^23 oxygen atoms

Molar Mass of H2O

We’ve been using the molecular weight of H2O interchangeably with molar mass, but these terms mean slightly different things:

Molecular weight – The sum of the atomic weights of each element in a molecule
Molar mass – The mass of 1 mole of a substance

So for H2O, the molecular weight is 2.016 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol.

The molar mass is the mass of 1 mole of H2O, which is also 18.02 g/mol. So the molar mass = molecular weight for H2O.

Molar Mass Calculation Example

Let’s look at calculating the molar mass of glucose (C6H12O6) as an example:

6 moles C x 12.011 g/mol = 72.066 g/mol

12 moles H x 1.008 g/mol = 12.096 g/mol
6 moles O x 16.00 g/mol = 96.00 g/mol

Adding these together:

Molar Mass of C6H12O6 = 72.066 + 12.096 + 96.00 = 180.16 g/mol

Molarity

Molarity is a common unit used in chemistry that measures the molar concentration of a solution. It is defined as the number of moles of solute dissolved per liter of solution.

The formula for calculating molarity is:

Molarity (M) = Moles solute / Liters solution

For example, to make a 1 M solution of HCl, you would need to dissolve 1 mole of HCl in enough water to make 1 liter of solution. Some examples:

1 M HCl – 1 mole HCl dissolved in 1 liter solution

0.5 M NaOH – 0.5 moles NaOH dissolved in 1 liter solution
2 M CuSO4 – 2 moles CuSO4 dissolved in 1 liter solution

Dilution of Solutions

The molarity of a solution can be decreased by adding more solvent, a process called dilution. For example, if you have 1 liter of a 2 M NaOH solution, you can dilute this to 1 M NaOH by adding enough water to make the volume 2 liters. The moles of NaOH stay the same, but the volume increases so the molarity decreases.

The dilution equation is:

M1 x V1 = M2 x V2

M1 = initial molarity

V1 = initial volume
M2 = final molarity
V2 = final volume

Empirical Formula

The empirical formula of a compound gives the simplest whole number ratio of atoms in that compound. Steps to determine the empirical formula:

Determine the mass composition from percent composition
Assume 100g total mass to determine moles of each element

Divide moles of each element by the smallest number to give whole number empirical formula

For example, a compound contains 40% C, 6.7% H, and 53.3% O. To find the empirical formula:

For 100 g total mass, 40 g is C, 6.7 g is H, and 53.3 g is O

Convert to moles: C = 40/12 = 3.3 mol, H = 6.7/1 = 6.7 mol, O = 53.3/16 = 3.3 mol
Divide each by the smallest mole amount: C = 3.3/3.3 = 1 mol, H = 6.7/3.3 = 2 mol, O = 3.3/3.3 = 1 mol

Therefore the empirical formula is CH2O.

Limiting Reactant and Reaction Yield

When reactants are mixed together, one reactant will limit how much product can be formed if it runs out first. This is called the limiting reactant.

For example:

2 H2 + O2 → 2 H2O

If you mix 2 moles of H2 with 1 mole of O2, O2 is the limiting reactant. 1 mole of O2 will produce 2 moles of H2O, leaving 1 mole of excess H2.

The reaction yield is the theoretical amount of product formed from the limiting reactant. Dividing the actual yield by the theoretical yield gives the percent yield.

For example, if the reaction above produced 1.5 moles of H2O, the percent yield would be (1.5/2) x 100% = 75%.

Being able to determine reaction yields and the limiting reactant is important for maximizing the efficiency of chemical production.

Oxidation Numbers

Oxidation numbers are used to keep track of electron transfer in oxidation-reduction reactions. Rules for assigning oxidation numbers:

Elements have an oxidation number of 0

Monoatomic ions have an oxidation number equal to their charge
Oxygen usually has an oxidation number of -2
Hydrogen has an oxidation number of +1 when bonded to non-metals

Total oxidation number of a neutral compound is 0

For example, in the compound NaCl:

Na = +1

Cl = -1
Sum = 0 (neutral compound)

Being able to assign oxidation numbers allows you to identify what is being oxidized and reduced in redox reactions.

Balancing Redox Reactions

Redox reactions involve simultaneous electron transfer – one species is oxidized while another is reduced. An example reaction:

Zn + CuSO4 → ZnSO4 + Cu

Balancing redox reactions has two steps:

Balance elements other than O and H
Balance oxygen by adding H2O, balance hydrogen by adding H+

So the balanced reaction is: