Step 1. Calculate the molar mass of the compound
To do this, we can use the Ideal Gas Law:
#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#
Since #n = m/M#. we can write
#pV = m/MRT#
And we can rearrange this to
#M = (mRT)/(pV)#
STP is defined as 0 °C and 1 bar, so
#M = ("16.96 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 1 color(red)(cancel(color(black)("L")))) = "385.16 g·mol"^"-1"#
Step 2. Calculate the formula of the compound
Your compound contains 37.321 % #"Cl"#.
(a) Calculate the mass of #"Cl"# in 1 mol of the compound.
#"Mass of Cl" = 385.16 color(red)(cancel(color(black)("g compound"))) × "37.321 g Cl"/(100 color(red)(cancel(color(black)("g compound")))) = "143.74 g Cl"#
(b) Calculate the moles of #"Cl"#
#"Moles of Cl" = 143.74 color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)("g Cl")))) = "4.055 mol Cl" ≈ "4 mol Cl"#
∴ The formula of the compound is #"MCl"_4#.
