I don't understand how to start my chemistry problem

Question

okay, so I might be doing something wrong, but I'm not even sure where to start with the 1st problem, but I worked out the second one and just need to know if I got the right answer for it(and what sig-fig to round it to).

For this one I think i'm supposed to find P in the PV=nRT set up, but I also need n and R and I can't have R with psi but I don't have any numbers to change from psi, if that makes sense.

1. A 38.3 mg sample of butane (C₄H₁₀) gas at 258.7 K occupies a volume of 4.09 mL, what is the pressure in psi?

For this one I got 1.40067 without rounding it to any sig-figs because i'm not sure how many i'd round it to.

1. What is the mass in grams of a 1.4 L sample of nitrogen gas at -16.1^oC when the pressure is 600.7 mm Hg?

Thank you so much for your help!

Answer 1

You're essentially correct in your approach, and all you need is the conversion factor for psi and atm. It happens to be 1 psi is about equal to 0.068 atm or put another way, 1 atm = 14.7 psi

C4H10 molar mass = 4x12 + 1x10 = 48 + 10 = 58 g/mol

moles present = 38.3 mg x 1 g/1000 mg x 1 mol/58 g = 6.60x10^-4 moles

Volume = 4.09 ml x 1 L/1000 ml = 0.00409 L

P = pressure = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 258.7K

PV = nRT

Solving for P we have...

P = nRT/V = (6.60x10^-4)(0.0821)(258.7) / 0.00409

P = 4.09x10^-7 atm

Converting to psi...

4.09x10^-7 atm x 1 psi/0.068 atm = 6.01x10^-6 psi (to 3 sig. figs.)

(as always, be sure to check the math)

V = 1.4 L

P = 600.7 mm Hg x 1 atm/760 mm Hg = 0.7904 atm

T = -16.1ºC + 273 = 256.9K

molar mass N₂ = 28 g/mol

PV = nRT

solve for n..

n = PV/RT = (0.7904 atm)(1.4 L) / (0.0821 Latm/Kmol)(256.9K)

n = 0.0525 moles

mass = 0.0525 moles N₂ x 28 g/mol = 1.47 g = 1.5 g (to 2 sig. figs.)