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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: pizza1512 on August 01, 2007, 07:28:30 AM

Title: Small calculation question
Post by: pizza1512 on August 01, 2007, 07:28:30 AM
I've forgotten how to do this sort of question so can anyone try and help me out?  I used to have a textbook that showed how to work out these problems but I had to return it back to school!  :-[

A sample of hydrated copper (II) sulphate (CuSO4.xH2O) weighing 0.869g was heated to constant mass and the residue weighed 0.556g.  Find the value of x.

Thanks
 ::)
Title: Re: Small calculation question
Post by: kylon on August 01, 2007, 09:04:48 AM
0.556g is constant mass for CuSO4, then the mole of CuSO4 is 0.556/160 = 0.003475 mol

then find the mass of H2O,
0.869g - 0.556g = 0.313g
then mole of H2O is 0.313/18 = 0.0174

from (CuSO4.xH2O), we know CuSO4 is 1 mol, 0.003475/0.003475 = 1

then, x=0.0174/0.003475 = 5
x = 5
give me a snack
Title: Re: Small calculation question
Post by: kylon on August 01, 2007, 09:06:02 AM
May i know how to delete posted thing??
Sorry for posted 2 same thing to u
i can only modify the text... :P
Title: Re: Small calculation question
Post by: profmsg on August 01, 2007, 10:28:49 AM
the answer from kylon is correct... :)
Title: Re: Small calculation question
Post by: pizza1512 on August 01, 2007, 02:37:19 PM
Ahhh.  I've forgotten another how to do another small calculation:

Calculate the volume of oxygen needed to burn 250 cm3 of butane and the mass of water formed.

Thanks again.
Title: Re: Small calculation question
Post by: Borek on August 01, 2007, 02:41:01 PM
Start with the reaction equation. Convert 250 mL of gas to moles (pV=nRT, or - as p not T are not given - assume molar volume 22.4L). Then it is simple stoichiometry (http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations).
Title: Re: Small calculation question
Post by: pizza1512 on August 01, 2007, 05:40:51 PM
Quote from: pizza1512 on August 01, 2007, 02:37:19 PM
Ahhh.  I've forgotten another how to do another small calculation:

Calculate the volume of oxygen needed to burn 250 cm3 of butane and the mass of water formed.

Thanks again.

I've written the equation for the reaction:

2C4H10 (g) + 13O2 (g) -> 8CO2 (g) + 10H2O (l)

But I can't remember how to calculate the ratios of gases to each another and how to work out moles, and masses etc.... If I don't reconquer this sort of question I will probably have to give up chemistry  :'( So please do help me.
Title: Re: Small calculation question
Post by: sdekivit on August 01, 2007, 06:06:55 PM
well you know your ratios:

2 mol butane reacts with 13 mol oxygen to yield 8 mol carbon dioxide and 10 mol water.

2 : 13 : 8 : 10

Whem for example 3 mol butane reacts, then you need (13 * 3)/2 = 19.5 mol oxygen. The yield will then be (8 * 3)/2 = 12 mol CO2 and (10 * 3) / 2 = 15 mol H2O.
Title: Re: Small calculation question
Post by: PanCerowany on March 19, 2010, 10:14:39 AM
Quote from: pizza1512 on August 01, 2007, 05:40:51 PM
I've written the equation for the reaction:

2C4H10 (g) + 13O2 (g) -> 8CO2 (g) + 10H2O (l)

But I can't remember how to calculate the ratios of gases to each another and how to work out moles, and masses etc.... If I don't reconquer this sort of question I will probably have to give up chemistry  :'( So please do help me.

Ratios are expplicitely given by the reaction equation.