Molar mass of substance A in material Eu(A)3 is 238g/mol. 0.476g of substance A was burned and produced 1.408g of CO2 and 0.252g of H2O. Substance A is made up of carbon, hydrogen and oxygen. What is the chemical formula for substance A?

1 Answer
Jun 21, 2016

Answer:

#"C"_16"H"_14"O"_2#

Explanation:

The way I see it, the part about how

The molar mass of substance #"A"# in material #"Eu"("A")_3# is #"238 g mol"^(-1)#

does not make sense because the molar mass of a substance, which represents the mass of one mole of said substance, is constant. It could be that the problem meant something else entirely, but I'm not sure what that something else could be.

So my instinct would be to take the info given by the problem and say that the molar mass of substance #"A"# is #"238 g mol"^(-1)#.

Even if that is not the correct interpretation, you can use these techniques to find the molecular formula of substance #"A"# in any given context.

Now, the idea here is that all the carbon that was initially present in the sample is now present in the carbon dioxide, #"CO"_2#. Since #1# mole of carbon dioxide contains #1# mole of carbon, you can say that the sample contained

#1.408 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole CO"_2))))/(44.01color(red)(cancel(color(black)("g")))) * (1color(red)(cancel(color(black)("mole C"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "0.38426 g C"#

Likewise, all the hydrogen that was initially present in the sample is now present in the water, #"H"_2"O"#. Since #1# mole of water contains #color(blue)(2)# moles of hydrogen, you can say that the sample contained

#0.252 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/(18.015color(red)(cancel(color(black)("g")))) * (color(blue)(2)color(white)(a)"moles H")/(1color(red)(cancel(color(black)("mole H"_2"O")))) * "1.00794 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.028120 g H"#

This means that the sample of substance #"A"# also contained

#m_"A" = m_"C" + m_"O" + m_"H"#

#m_"O" = "0.476 g" - ("0.38426 g" + "0.028120 g")#

#m_"O" = "0.063620 g O"#

At this point, you must figure out how many moles of each element you have in that sample. This will allow you to calculate the empirical formula of substance #"A"#.

You will have

#"For C: " 0.38426 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.031992 moles C"#

#"For H: " 0.028120 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.027898 moles H"#

#"For O: " 0.063620 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994 color(red)(cancel(color(black)("g")))) = "0.0039764 moles O"#

To find the mole ratios that exist between the three elements in the compound, divide all values by the smallest one

#"For C: " (0.031992 color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 8.045 ~~ 8#

#"For H: " (0.027898color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 7.016 ~~ 7#

#"For O: " (0.0039764color(red)(cancel(color(black)("moles"))))/(0.0039764color(red)(cancel(color(black)("moles")))) = 1#

Since #8:7:1# is the smallest whole number ratio that can exist between the three elements, it follows that the empirical formula for substance #"A"# is

#"C"_8"H"_7"O"_1 -># empirical formula

Now, if you take #"238 g mol"^(-1)# to be the molar mass of substance #"A"#, you can say that you must have

#(8 xx 12.011 color(red)(cancel(color(black)("g mol"^(-1)))) + 7 xx 1.00794color(red)(cancel(color(black)("g mol"^(-1)))) + 1 xx 15.9994color(red)(cancel(color(black)("g mol"^(-1))))) xx color(blue)(n) = 238color(red)(cancel(color(black)("g mol"^(-1))))#

#119.43 * color(blue)(n) = 238 implies color(blue)(n) = 238/119.143 = 1.9976 ~~ 2#

Therefore, the molecular formula for substance #"A"# is

#("C"_8"H"_7"O"_1)_color(blue)(2) = color(green)(|bar(ul(color(white)(a/a)color(black)("C"_16"H"_14"O"_2)color(white)(a/a)|))) -># molecular formula