# Homework Help: What is this element, carbon, hydrogen, oxyden or aluminum?

1. Jan 20, 2004

### Rival

8. [CJ6 14.P.001.] A mass of 135 g of an element is known to contain 30.1 1023 atoms. What is the element?
carbon
hydrogen
oxyden
aluminum

This is the question on my Physics practice test for my exam tomorrow. i don't know where to even begin with this one. we haven't really gone over anything like this in class...

Kevin

2. Jan 20, 2004

### himanshu121

Can u find the no. of moles

3. Jan 20, 2004

### Rival

how?

4. Jan 20, 2004

### himanshu121

1 mole contains 6.02 * 1023 entities here atom

5. Jan 20, 2004

### Rival

ok so i am guessing that the equation that i need to use is n=N/NA

but how do we figure out N?

6. Jan 20, 2004

### himanshu121

yup, N is given to u 30.1 1023

7. Jan 20, 2004

### Rival

ok so we have 5 moles but now what? how do we figure out what element it is?

8. Jan 20, 2004

### himanshu121

Do u now approx mol wt of above given elements

Find the Mol wt with the help of Mole equation

$$n=\frac{g}{M_0}$$ where M0 is Mol wt and g is the given wt

9. Jan 20, 2004

### himanshu121

so i feel u got M0=135/5=27

Which belongs to ????(WHAT)

10. Jan 20, 2004

### Rival

so using the equation you gave me, i have n = 5 and g = 135, so i find Mo, which came out to be 27, and in the periodic table 27 is Co. what element is Co? but i think its wrong...

11. Jan 20, 2004

### Rival

now that i know the mol wt, i matched it with the periodic table and it comes out to Co, but which one of the choices are Co.

12. Jan 21, 2004

### himanshu121

U checked in for Atomic Number Rather than atomic wt

Atomic wt for Al is 27 (Aluminum)

Co is Cobalt

13. Jan 21, 2004

### Rival

hehe, that could help if i looked for the right thing. Thanks alot, i really appreciate your help.

Kevin

14. Jan 21, 2004

### HallsofIvy

The whole point of a "mole" of an element is that it's weight in grams is equal to the weight of an individual atom in "atomic units". Once you know that you can identify the element.