Two unknwon factors in this stoichiometry problem

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To calculate the volume occupied by 36.25g of C4H10 at 85.20 kPa and 108°C, the ideal gas law (PV = nRT) is applied. The volume (V) can be isolated as V = nRT/P. To find the number of moles (n), the mass is divided by the molar weight of C4H10. The molar weight is calculated by summing the atomic weights of its components: 4 carbons (12.01 g/mol each) and 10 hydrogens (1.01 g/mol each), totaling 58.14 g/mol. Thus, 36.25g of C4H10 corresponds to approximately 0.6234 moles. It is confirmed that under the given conditions, butane is in a gaseous state, making the ideal gas law applicable.
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Can someone help me with this?

What volume, cm³, would 36.25g of
C_4H_1_0
occupy at 85.20 kPa and 108 celsius?

So I am assuming i should use the ideal gas equation, PV = nRT, and isolate V so i get:

V=\frac{nRT}{P}

heres where I am stuck, how do i solve for "n" when only the mass is given? how do i find the molecular weight?

and its my first time here .. so can some1 explain to me how these latex things work =X
 
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you can find the number of moles (n) by dividing the number af grams you know you have by the molar weight of the substance.

yuo can find the molar weight of the substance(grams/mole) by adding up each elements molar weight and totallying them together,

for example:
C4H10 is, 4 carbons (12.01 g/mol each) + 10 hydrogens (1.01) g/mol each) = 48.04 + 10.1 = 58.14 g/mol


and since you already know you have 36.25 g of it, that means you have,
36.25/58.14 = .6234 moles of it.


remember, the ideal gas law only works for gasses, so make sure that your butane (C4H10) is a gas under those conditions, otherwise it won't work, but I am pretty sure that at that low of a pressure and that high of a temperature, it would be.
 
alright tahnks i understand now
 
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