Question #765f8

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Question #765f8

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Answer 1 · 2015-01-25T19:51:01

The empirial formulae for the 2nd and 3rd compounds are $X_{3} Z$ $X_3Z$ and $X Z_{4}$ $XZ_4$ .

For the 1st compound:

Ratio by mass X : Z = 1 : 0.472

Ratio by moles = 1/MrX : 0.472/MrZ = 2 :3

This means $\frac{\frac{I}{M r X}}{\frac{0.472}{M r Z}} = \frac{2}{3}$ $((I)/(MrX))/((0.472)/(MrZ))= 2/3$

So

$\frac{1}{M r X} \times \frac{M r Z}{0.472} = \frac{2}{3}$ $(1)/(MrX)xx(MrZ)/(0.472)=2/3$

So

$\frac{M r Z}{M r X} = \frac{2 \times 0.472}{3} = 0.314$ $(MrZ)/(MrX)= (2xx0.472)/(3)=0.314$

So

$M r Z = 0.314 M r X$ $MrZ=0.314MrX$

For the 2nd compound ratio X : Z by moles = $\frac{1}{M r X} : \frac{0.105}{M r Z}$ $(1)/(MrX):(0.105)/(MrZ)$

We can substitute $M r Z$ $MrZ$ for $0.314 M r X$ $0.314MrX$ ---->

Ratio by moles = $\frac{1}{M r X} : \frac{0.105}{0.314 M r X}$ $(1)/(MrX):(0.105)/(0.314MrX)$

= $\frac{1}{M r X} : \frac{1}{3 M r X}$ $(1)/(MrX):(1)/(3MrX)$

So the mole ratio X : Z = 1 : 1/3 = 3 : 1

So empirical formula is $X_{3} Z$ $X_3Z$

For the 3rd compound ratio in moles X : Z= $\frac{1}{M r X} : \frac{1.259}{0.314 M r X}$ $(1)/(MrX):(1.259)/(0.314MrX)$

= 1 :4

So the empirical formula is $X Z_{4}$ $XZ_4$

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Answer 2 · 2015-01-25T20:06:34

So, you know that you have three reactions that produce three different compounds; however, all three reactions use the same amount of $X$ $"X"$ , which means that you can use this to determine the formulas of the other two compounds.

Here's you generic reaction:

$a X + b Z \to X_{a} Z_{b}$ $a"X" + b"Z" -> "X"_a"Z"_b$

For the first reaction:

$2 X + 3 Z \to X_{2} Z_{3}$ $2"X" + 3"Z" -> "X"_2"Z"_3$

Notice that the balanced chemical equation has a $2 : 3$ $2:3$ mole ratio between $X$ $"X"$ and $Z$ $"Z"$ , which means that 2 moles of the former will need 3 moles of the latter in order to produce the compound.

In other words, $n_{X} = 2/3 \cdot n_{Z}$ $n_("X") = "2/3" * n_("Z")$ . Now, let's assume the molar mass of $X$ $"X"$ is $M_{X}$ $M_("X")$ and the molar mass of $Z$ $"Z"$ is $M_{Z}$ $M_("Z")$ . The above equation can be written as

$n_{X} = 2/3 \cdot n_{Z} \Rightarrow \frac{1.00 g X}{M_{X}} = \frac{2}{3} \cdot \frac{0.472 g Z}{M_{Z}}$ $n_("X") = "2/3" * n_("Z") => ("1.00 g X")/(M_("X")) = 2/3 * ("0.472 g Z")/(M_("Z"))$

This will get you

$1.00 g X \cdot \frac{M_{X}}{M_{Z}} = \frac{2}{3} \cdot 0.472 g Z = 0.315 g = constant$ $"1.00 g X" * (M_("X"))/(M_("Z")) = 2/3 * "0.472 g Z" = "0.315 g" = "constant"$

SInce you'll always have 1.00 g of $X$ $"X"$ , and since the molar masses of the two elements are constant, the expression on the left will be constant. What this implies is that the product of the moles ratio and the mass of $Z$ $"Z"$ will be constant.

In general form,

$1.00 g X \cdot \frac{M_{X}}{M_{Z}} = \frac{a}{b} \cdot m_{Z}$ $"1.00 g X" * (M_("X"))/(M_("Z")) = a/b * m_("Z")$ , or $\frac{a}{b} \cdot m_{Z} = 0.315 g$ $a/b * m_("Z") = "0.315 g"$

Let's solve for $m_{Z} = 0.105 g$ $m_("Z") = "0.105 g"$

$\frac{a}{b} \cdot 0.105 g = 0.315 g \Rightarrow \frac{a}{b} = \frac{0.315 g}{0.105 g} = 3$ $a/b * "0.105 g" = "0.315 g" => a/b = ("0.315 g")/("0.105 g") = 3$

This means the second formula is $X_{3} Z$ $"X"_3"Z"$ .

For $m_{Z} = 1.259 g$ $m_("Z") = "1.259 g"$ , you'll get $\frac{a}{b} = \frac{0.315 g}{1.259 g} = 0.25 = \frac{1}{4}$ $a/b = ("0.315 g")/("1.259 g") = 0.25 = 1/4$ .

The third formula is ${XZ}_{4}$ $"XZ"_4$ .

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