Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109	General Chemistry	Spring 2008
Lecture Notes:: Mole Problems	© R. Paselk 2001
	Answers
	Stoichiometry - Mole Problems

Stoichiometry: Moles, Atoms, & Formulae Problems

• How many atoms are there in 1.00 g of ethanol?

First need formula = C₂H₅OH

so weight = (2 [12.01]g/mole + 6 x [1.008 g/mole] + 16.00 g/mole) = 46.08 g/mole

then find moles = (1.00 g)/(46.08 g/mole) = 2.170 x10^-2mole

then multiply by Avogadro's Number: (2.170 x10^-2mole) x (6.022 x 10^-3 atoms/mole) = 1.307 x10²² glucose molecules

But - there are 10 atoms/formula, So 1.31 x10²³ atoms

 

 

• How many atoms are there in 1.00 g of glucose?

First need formula = C₆H₁₂O₆

so MW = (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 180.16 g/mole

then find moles = (1.00 g)/(180.16 g/mole) = 5.551 x10^-3mole

then multiply by Avogadro's Number: (5.551 x10^-3mole) x (6.022 x 10^-3 atoms/mole) = 3.343 x10²¹ glucose molecules

But - there are 24 atoms/formula, So 8.02 x10²² atoms

 

• How many moles are there in 23.5 g of propane (C₃H₈)

and moles = (23.5 g)/(45.102 g/mole) = 5.21 x 10^-1mole.

• How many moles are there in 23.5 g of NaCl

and moles = (23.5 g)/(58.44g/mole) = 4.021 x 10^-1mole = 4.021 x 10^-1mole

 

 

 

 

 

 

 

• ethanol? (C₂H₅OH)

Syllabus / Schedule

C109 Home

C109 Lecture Notes

© R A Paselk

Last modified 31 January 2007