## Fundamentals E17

Rida Ismail 2E
Posts: 139
Joined: Sat Sep 07, 2019 12:16 am

### Fundamentals E17

does anyone know how to complete Fundamentals E17?

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

### Re: Fundamentals E17

The question is asking for the sample with more moles. In part a, you would convert both the 75g indium and 80g tellurium into their respective moles by dividing the substance molar mass.
$75g \textrm{In} \div 114.82g/mol = 0.65 mol \textrm{In}$
$80g \textrm{Te} \div 127.60g/mol = 0.63 mol \textrm{Te}$
0.65 > 0.63, so the 75g of indium contains more moles of atoms than the 80g of tellurium. Then do the same for part b. But for part c, you will have to divide by Avogadro's number ($6.022\times 10^{23}\textrm{atoms/mol}$) to convert atoms into moles. But since there are the same number of rubidium and iron atoms, there are the same amount of rubidium and iron moles.

Jenna Ortiguerra 4G
Posts: 50
Joined: Sat Aug 17, 2019 12:18 am

### Re: Fundamentals E17

In a and b, convert the following given grams to moles by using the element's molar mass. In c, convert the following given atoms to moles by using Avogadro's number (6.022*10^23)
a. 75g/114.82 g*mol^-1 = 0.65 mol In
80g/127.60 g*mol^-1 = 0.63 mol Te
0.65 mol In > 0.63 mol Te

b. 15.0g/30.97 g*mol^-1 = 0.484 mol P
15.0g/32.07 g*mol^-1 = 0.464 mol S
0.484 mol P > 0.468 mol S

c. 7.36 * 10^23 atoms Ru / 6.022*10^23 atoms*mol^-1) = 1.22 * 10^4 moles
7.36 * 10^23 atoms Fe / 6.022*10^23 atoms*mol^-1) = 1.22 * 10^4 moles
The two samples have the same number of moles because they have the same number of atoms.