Go Back   Chemistry Help Forum > Chemistry Forums > Undergraduate Chemistry Forum

Undergraduate Chemistry Forum Ask your undergraduate chemistry questions here.

Reply
 
Thread Tools Display Modes
Old Oct 2nd 2013, 06:41 PM   #1
AK16
Junior Member
 
Join Date: Oct 2013
Posts: 1
AK16 is on a distinguished road
Calculate the Mass Fractions of Gold and Copper

A 9.40g wedding band is made up of gold and copper, and has a 4.382x10^22 atoms. Calculate the mass fractions of gold and copper in this 18-carat gold alloy.

I have gotten this far:
1)
(Au) = gold = x
(Cu) = copper = y

9.40g = x + y

2) Ntotal = 4.382x10^22

4.382x10^22 = N(Au) + N(Cu)

3) Divide by Avogadro Number

4.382x10^22/6.022x10^23 = N(Au)/Na + N(Cu)/Na

7.277x10^-2 mol = n(Au) + n(Cu)

4) Find Masses using mol amount

7.277 x 10^-2 mol = m(Au)/M(Au) + m(Cu)/M(Cu)
7.277 x 10^-2 mol = x/(196.97g/mol) + (9.40g-x)/(63.55g/mol)

5) solve for x

This is where I am stuck, how do I solve for X?
AK16 is offline   Reply With Quote
Old Oct 2nd 2013, 07:26 PM   #2
bjhopper
Senior Member
 
Join Date: May 2010
Posts: 491
bjhopper is on a distinguished road
Originally Posted by AK16 View Post
A 9.40g wedding band is made up of gold and copper, and has a 4.382x10^22 atoms. Calculate the mass fractions of gold and copper in this 18-carat gold alloy.

I have gotten this far:
1)
(Au) = gold = x
(Cu) = copper = y

9.40g = x + y

2) Ntotal = 4.382x10^22

4.382x10^22 = N(Au) + N(C

3) Divide by Avogadro Number

4.382x10^22/6.022x10^23 = N(Au)/Na + N(Cu)/Na

7.277x10^-2 mol = n(Au) + n(Cu)

4) Find Masses using mol amount

7.277 x 10^-2 mol = m(Au)/M(Au) + m(Cu)/M(Cu)
7.277 x 10^-2 mol = x/(196.97g/mol) + (9.40g-x)/(63.55g/mol)

5) solve for x

This is where I am stuck, how do I solve for X?


An 18 caret gold alloy contains 75% gold by weight
bjhopper is offline   Reply With Quote
Reply

  Chemistry Help Forum > Chemistry Forums > Undergraduate Chemistry Forum



Thread Tools
Display Modes



Facebook Twitter Google+ RSS Feed