Question

purpose (lo): use water displacement and the pv=nrt equation to determine the molar mass of a gas
safety concerns: butane is highly flammable, keep flames and sparks away from the butane gas
necessary ppe: goggles
methodology:
method/lab skill: water displacement & gas collection
procedure: must be detailed
raw data:

mass of lighter after dried	22.22	g
mass of lighter after gas released (dried again)	21.71	g
mass of butane collected	0	g
volume of gas collected	75	ml
converted to liters	0.075	l
temperature of water basin	20	°c
converted to kelvin	293	k
pressure in the room (p_total)	758	mm hg
partial pressure of water vapor (p_water)	mm hg
partial pressure of c₄h₁₀ (p_butane)	mm hg
convert to atm	0.96	atm
the correct r constant	0.0820
temperature (°c)	p_water vapor (mm hg)
16	13.5
17	14.5
18	15.5
19	16.5
20	17.5
21	18.7
22	19.3
23	21.1
24	22.4
25	23.8
26	25.2
27	26.7
28	28.3
29	30.0

Explanation:

Step1: Calculate partial pressure of butane

The total pressure in the room $P_{total}=759$ mmHg. The temperature of water basin is $20^{\circ}C$, and from the table, the partial - pressure of water vapor $P_{H_2O}=17.5$ mmHg. The partial pressure of butane $P_{C_4H_{10}}=P_{total}-P_{H_2O}$.
$P_{C_4H_{10}} = 759 - 17.5=741.5$ mmHg. Convert this to atm: $P_{C_4H_{10}}=\frac{741.5}{760}\approx0.976$ atm.

Step2: Use the ideal gas law $PV = nRT$ to find the number of moles of butane

We know that $P = 0.976$ atm, $V=0.075$ L, $T = 293$ K, and $R = 0.0821$ L·atm/(mol·K).
From $PV=nRT$, we can solve for $n$: $n=\frac{PV}{RT}$.
$n=\frac{0.976\times0.075}{0.0821\times293}$
$n=\frac{0.0732}{24.0553}\approx0.00304$ mol.

Step3: Calculate the molar mass of butane

The mass of butane collected $m = 0.5$ g (assuming the difference between the mass of the lighter before and after gas release is 0.5 g, as the value in the table for mass of butane collected seems unclear, if we take $22.21 - 21.71$).
The molar mass $M=\frac{m}{n}$.
$M=\frac{0.5}{0.00304}\approx164.47$ g/mol. But this is likely due to incorrect or unclear mass - value input. If we assume the correct mass of butane is based on proper subtraction and accurate data, the molar mass of butane ($C_4H_{10}$) from its formula is $4\times12 + 10\times1=58$ g/mol.

Answer:

If we assume the correct mass data and calculations, the molar mass of butane ($C_4H_{10}$) is 58 g/mol. If we calculate based on the data in the table with some assumptions for the mass value, we get approximately 164.47 g/mol which is likely incorrect due to data issues.