Question

29. A 1.58-g sample of C₂H3X3 (9) has a volume of 297 mL at 769 mm Hg and 35°C. Identify the element X

Answer 1

First you convert mmHG (weird unit of pressure measurement) into atmospheres
1atmosphere is 760 mmHG
take 769 and divide it by 760 and you get 1.0118 atm
let's make that into 1.012 atm
here we can use PV=nRT
P=Pressure V=Volume n=number of moles R=gas constant and T=Temperature
P=1.012atm V=0.297L n=? R=0.0821 atm•L/mol T=35+273.15
PV=nRT rearranged is also PV/RT = n
1.012 x 0.297 / \[\frac{1.012*0.297}{0.0821*(35+273.15)}=0.01188041558\]

Answer 2

now that we have the number of moles we can use the equation n = m/Mr
n=number of moles m=mass Mr = molecular mass
\[Mr=\frac{ 1.58g }{0.012}\]
Mr=131.6666667
Your compound is C₂H3X3
Carbon has an atomic mass of 12
Hydrogen has an atomic mass of 1
2 Carbons = 24 and 3 Hydrogens = 3
\[131.6666667-24-3 =104.6666667\]
X3=104.6666667
To find 1 X you divide it by 3 and you get
\[\frac{ 104.6666667 }{ 3 } =34.8888889\]
Rounded up X = 35 which is the mass of Cl Chlorine