Mole Conversions

What is the molar mass of a substance?

• The mass of one atom of that substance.

• The mass of one mole of that substance in grams.

• The mass of one mole of a substance in atomic mass units (amu).

• The mass of a substance in mg.

Where can you find the molar mass of a substance?

• Looking up the average atomic mass on a periodic table.

• Looking up the atomic number on a periodic table.

• Looking up the conversion factor for one mole which is 1 mole = 6.022²³ items.

• Looking up the metric prefixes chart.

What are the units of molar mass?

• mol

• g

• g/mol

• mol/g

• amu

What is the molar mass of carbon?

• 6 amu

• 6 g/mol

• 12.01 amu

• 12.01 g/mol

What is the molar mass of MgCl₂?

• 24.31 g/mol

• 59.76 g/mol

• 95.21 g/mol

• 119.52 g/mol

(Hint: write out how you would solve the problem first on paper before choosing an answer.) 

How would you set up the problem if you have a sample of 2.50 g of sodium (Na), how many moles of Na do you have?

•  $2.50g\ Na\ \times\frac{22.99\ mol\ Na}{1\ g\ Na}=$  

•  $2.50\ g\ Na\ \times\frac{1\ mol\ Na}{22.99\ g\ Na}=$  

•  $2.50\ g\ Na\ \times\frac{22.99\ g\ Na}{1\ mol\ Na}=$  

•  $2.50\ g\ Na\ \times\frac{1\ g\ Na\ }{22.99\ mol\ Na}=$  

How would you set up the problem if you want to find the mass, in grams, of 0.55 moles of silicon?

•  $0.55\ mol\ Si\ \times\ \frac{28.09\ g\ Si}{1\ mol\ Si}\ =\$  

•  $0.55\ mol\ Si\ \times\frac{1\ g\ Si\ }{28.09\ mol\ Si}=$  

•  $0.55\ mol\ Si\ \times\frac{28.09\ mol\ Si}{1\ g\ Si}=$  

•  $0.55\ mol\ Si\ \times\frac{1\ mol\ Si}{28.09\ g\ Si}=$  

Tags

NGSS.HS-PS1-7

What is the mass, in grams, of 3.33 mol of aluminum?

• 89.8 g Al

• 90. g Al

• 0.123 g Al

• 0.12 g Al

How many moles are in a sample of 145.0 g of KBr?

• 17,255 mol KBr

• 17,260 mol KBr

• 1.22 mol KBr

• 1.218 mol KBr

How would you set up the problem if you have 3.80 moles of calcium (Ca), and you are trying to determine how many atoms of calcium you have?

•  $3.80\ mol\ Ca\ \times\frac{40.08\ g\ Ca}{1\ mol\ Ca}=$  

•  $3.80\ mol\ Ca\ \times\frac{6.022\ x\ 10^{23\ }atoms\ Ca}{1\ mol\ Ca}=$  

•  $3.80\ mol\ Ca\ \times\frac{1\ g\ Ca}{40.08\ mol\ Ca}=$  

•  $3.80\ mol\ Ca\ \times\frac{1\ atoms\ Ca}{6.022\ x\ 10^{23}\ mol\ Ca}=$  

Tags

NGSS.HS-PS1-7