Answer
1.418 g is the mass of chlorine that is combined with 4.315 g of metallic element X.
atomic mass of X= 107.88 amu
We see in the periodic table of elements that Ag ( silver) is the element with the atomic mass 107.8682 amu. The atomic mass that we found is that of Ag ( within experimental error).
Work Step by Step
Based on the chemical formula of XCl, note that XCl contains one atom X for every one atom Cl.
1- First we are required ti find the mass of chlorine that is combined with metallic element X.
We are given the volume and density of chlorine that is used in this chemical reaction. We can find the mass of chlorine .
We know: d = $\frac{m}{v}$
By multiplying both sides with v we get an identical equation with this one: d x v = $\frac{m}{v}$ x v
m = d x v
Replace density and volume to find the mass of chlorine :
m = 0.4810L x 2.948 $\frac{g}{L}$
m = 1.4180 g
So we found out that 1.418 g is the mass of chlorine that is combined with 4.315 g of metallic element X.
2- We have to find atomic mass of X.
Based on the chemical formula of XCl, we notice that XCl contains one atom X for every one atom Cl,
so in XCl there is this ratio:
1 atom X : 1 atom Cl
atomic mass X : atomic mass Cl
replace atomic mass of Cl:
atomic mass X : 35.453 amu Cl
We know as well that 4.315 g of X is combined with 1.418 g of chlorine in XCl.
So we have these ratios:
atomic mass X : 35.453 amu Cl
4.315 g X : 1.418 g Cl
By applying mathematical rule of three we can find atomic mass of X:
atomic mass of X = $\frac{4.315 g * 35.453 amu}{1.418 g}$
atomic mass of X= 107.884 amu
atomic mass of X= 107.88 amu
We see in the periodic table of elements that Ag ( silver) is the element with the atomic mass 107.8682 amu. The atomic mass that we found is that of Ag ( within experimental error).
