Question #77d3f

1 Answer
Feb 1, 2016

Answer:

#"40 g mol"^(-1)#

Explanation:

The interesting thing about this problem is that you don't really need to know the compound's molecular formula, but you do need to know its empirical formula.

As you know, a compound's empirical formula tells you the smallest whole number ratio that exists between the elements that make up said compound.

In this case, if the molecular formula is said to #"M"_4"O"_8#, the empirical formula will be #"M"_1"O"_2#, since #1:2# is the smallest whole number ratio that exists between #4# and #8#.

So, your compound contains two moles of oxygen for every one mole of #"M"#.

Now, a #"18.00-g"# sample of this compound is said to contain a total of #"10 g"# of #"M"#. This of course implies that it contains

#m_"compound" = m_"M" + m_"O"#

#m_"O" = "18.00 g" - "10 g" = "8 g oxygen"#

At this point, all you really need to know if how many moles of oxygen you have in the sample. Use oxygen's molar mass to help you with that

#8 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0 color(red)(cancel(color(black)("g")))) = "0.5 moles O"#

This means that sample must contain

#0.5 color(red)(cancel(color(black)("moles O"))) * "1 mole M"/(2color(red)(cancel(color(black)("moles O")))) = "0.25 moles M"#

Finally, to get the molar mass of element #"M#", divide its mass by the number of moles it contains

#M_M = "10 g"/"0.25 moles" = color(green)("40 g mol"^(-1))#