Question

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Answer 1

`\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}`

Step-by-step explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

• Na: 22.9897693 g/mol
• C: 12.011 g/mol
• O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

• Na₂ = 22.9897693 * 2= 45.9795386 g/mol
• O₃ = 15.999 * 3= 47.997 g/mol
• Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

`\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}`

We are converting 57.3 grams to moles, so we multiply by this value.

`57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}`

Flip the ratio so the units of grams of sodium carbonate cancel.

`57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}`

`57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }`

`\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3`

`0.5406295944 \ mol \ Na_2CO_3`

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

`0.541 \ mol \ Na_2CO_3`

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.