Now if your lab partner determined that there were 3.448 x 10(to the22 )atoms of O in an unknown sample of Mg(NO3)2, how many milligrams would the unknown sample weigh?
-
[3.448*10^22 atoms 0] * [1mol O/(6.02*10^23 atoms O)] * [1 mol Mg(NO3)2/6 mol O] *
[148.33g Mg(NO3)2/1mol Mg(NO3)2]
1.416g Mg(NO3)2 * 1000 =
1416mg Mg(NO3)2 or 1.416*10^3 mg Mg(NO3)2
The molar mass of Mg(NO3)2 is 24.31+2(14.01)+6(16)=148.33g Mg(NO3)2
Hope you can read all that.
[148.33g Mg(NO3)2/1mol Mg(NO3)2]
1.416g Mg(NO3)2 * 1000 =
1416mg Mg(NO3)2 or 1.416*10^3 mg Mg(NO3)2
The molar mass of Mg(NO3)2 is 24.31+2(14.01)+6(16)=148.33g Mg(NO3)2
Hope you can read all that.