# Question 8971d

Aug 24, 2016

For part (a)

${\text{CuC"_4"H"_4"O}}_{6}$

#### Explanation:

I'll show you how to approach part (a) and leave part (b) to you as practice.

So, the problem wants you to find the simplest formula for copper(II) tartrate. In other words, you must find the empirical formula of this compound.

To do that, you must find the smallest whole number ratio that exists between the elements that make up this compound.

The problem provides you with the percent composition of copper(II) tartrate. The first thing to do here is to convert the percentages by mass to grams.

To make the calculations easier, pick a $\text{100-g}$ sample of copper(II) tartrate. This sample will contain

• $\text{30.03% Cu " -> " 30.03 g Cu}$
• $\text{22.70% C " -> " 22.70 g C}$
• $\text{1.91% H " -> " 1.91 g H}$
• $\text{45.37% O " -> " 45.37 g O}$

To convert the grams to moles, use the molar masses of the four elements. You will have

$\text{For Cu: " 30.03 color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "0.4726 moles Cu}$

$\text{For C: " 22.70 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "1.890 moles C}$

$\text{For H: " 1.91 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "1.895 moles H}$

$\text{For O: " 45.37 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "2.836 moles O}$

Next, find the mole ratio that exists between the elements in the compound by dividing all values by the smallest one

"For Cu: " (0.4726 color(red)(cancel(color(black)("moles"))))/(0.4726 color(red)(cancel(color(black)("moles")))) = 1

"For C: " (1.890color(red)(cancel(color(black)("moles"))))/(0.4726 color(red)(cancel(color(black)("moles")))) = 3.999 ~~4

"For H: " (1.895color(red)(cancel(color(black)("moles"))))/(0.4726 color(red)(cancel(color(black)("moles")))) = 4.01 ~~4

"For O: " (2.836color(red)(cancel(color(black)("moles"))))/(0.4726 color(red)(cancel(color(black)("moles")))) = 6.001 ~~ 6#

This means that the four elements are present in a $1 : 4 : 4 : 6$ mole ratio in this compound. Since this is also the smallest whole number ratio that can exist here, you can say that the empirical formula for copper(II) tartrate will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Cu"_1"C"_4"H"_4"O"_6 implies "CuC"_4"H"_4"O}}_{6}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The strategy to use when dealing with similar problems is

• convert the percentages by mass to grams by picking a $\text{100-g}$ sample
• use the molar masses of the elements to convert the grams to moles
• divide all values by the smallest one to find the mole ratio that exist between the constituent elements
• make sure that this ratio is the smallest whole number ratio possible