How many formula units are in #"0.335 g CaO"#?

2 Answers
May 18, 2017

Answer:

#3.60 * 10^(21)#

Explanation:

Let's try to create a conversion factor that will take us directly from the number of grams of calcium oxide to the number of formula units present in the sample.

For starters, look up the molar mass of calcium oxide

#M_ "M CaO" = "56.0774 g mol"^(-1)#

This tells you that #1# mole of calcium oxide has a mass of #"56.0774 g"#, which can be written as a conversion factor

#color(blue)("1 mole CaO")/"56.0774 g"#

Now, you know that in order to have #1# mole of calcium oxide, you need to have #6.022 * 10^(23)# formula units of calcium oxide #-># this is known as Avogadro's constant.

This can be written as a conversion factor as well

#(6.022 * 10^(23)color(white)(.)"formula units")/color(blue)("1 mole CaO")#

As you can see, the two conversion factors have a common quantity. If you multiply these two conversion factors

#color(red)(cancel(color(blue)("1 mole CaO")))/"56.0774 g" * (6.022 * 10^(23)color(white)(.)"formula units")/color(red)(cancel(color(blue)("1 mole CaO")))#

you will get

#(6.022 * 10^(23)color(white)(.)"formula units")/"56.0774 g"#

You now have a conversion factor that takes you from grams to formula units or vice versa.

So, you know that your sample has a mass of #"0.335 g"#. Multiply this by the conversion factor to get

#0.335 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23)color(white)(.)"formula units")/(56.0774color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(3.60 * 10^(21)color(white)(.)"formula units")))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of calcium oxide.

May 18, 2017

Answer:

There are #3.60xx10^21color(white)(.)"formula units"# in #0.335 g CaO"#.

Explanation:

There are #6.022xx10^23# in 1 mole of anything, including formula units.

You need to determine the number of moles in #"0.335 g CaO"#. Once you know the number of moles of #"CaO"#, you can determine the number of formula units by multiplying the number of moles by #6.022xx10^23#.

#color(blue)("Number of Moles"#

You need to #color(red)("determine the molar mass"# of #"CaO"#, which is the sum of the atomic weights of each element on the periodic table in grams/mole, or g/mol.

#"Ca:"##"40.078 g/mol"#
#"O:"##"15.999 g/mol"#

#"Molar mass CaO"="40.078 g/mol + 15.999 g/mol"="56.077 g/mol"#

#color(red)("Determine the number of moles"# by multiplying the given mass of #"CaO"# by the inverse of its molar mass.

#0.335color(red)cancel(color(black)("g CaO"))xx(1"mol CaO")/(56.077color(red)cancel(color(black)("g CaO")))="0.00597 mol CaO"#

#color(blue)("Number of Formula Units"#

Multiply the mol #"CaO"# by #6.022xx10^23# formula units/mol.

#0.00597color(red)cancel(color(black)("mol CaO"))xx(6.022xx10^23"formula units CaO")/(1color(red)cancel(color(black)("mol CaO")))=3.60xx10^21color(white)(.)"formula units"# rounded to three significant figures