How do you do this chemistry problem?

Question

calculate the number of oxygen present in 50.0 grams of magnesium carbonate trihydrate

Answer 1

Oxygen atoms?

50.0g MgCO3.3H2O X 1mol/106.356g X 6molO/1mol cpd. X 6.022 X10^23 = 1.70 X 10^24 atoms O

Answer 2

Here goes little one...

Magnesium Carbonate Trihydrate = MgCO3·3H2O

In one molecule there are six oxygen atoms present ...
Get out a periodic table, sum the molecular weights of the atoms present and you'll find the molecular weight to be 138.787 g/mol and then:

50 grams *
(138.787 g/mol)^(-1) *
6.0221415 × 10^23 molecules / mole
*
6 oxygen atoms / molecule

and that's your answer :-)

Answer 3

I think that you want the weight of oxygen. Let magnesium carbonate trihydrate becalled MCT

Atomic weights: Mg=24 C=12 O=48 H=1 H2O=18 MgCO3*3H2O=138

There are six O's per molecule of MgCO3*3H2O

50.0gMCT x 1molMCT/138gMCT x 6molO/1molMCT x 16gO/1molO = 34.8g O

Answer 4

When you say "number of oxygen" I'll assume you mean the number of oxygen "molecules".

Magnesium Carbonate Trihydrate is MgCO3·3H2O

Therefore RAM of MgCO3·3H2O is 24+12+16x3+3(1x2+16) = 138g

Total Oxygen present in grams is 16x3+16x3 = 96g

Percentage present = (96/138) = 0.69565

So, 50 x 69.565% = 34.78g of O

Then to convert this to number of molecules...

34.78/(16x2) = 1.09 moles

1 moles =6x10^23

So, 6x10^23 x 1.09 = ? (I don't have a calculator)

Answer 5

MgCO3•3H2O
Mass is Mg=24.3
+C=12.0
+6 Oxygen=6(16.0)=96.0
+ 6 H=6(1.008)= 6.048
Mass is 138.3
Number of moles 50.0 g divided by 138.3g/mole= 0.362 moles
Since there are 6 oxygens in each molecule, there would be 6 X 0.362=2.17moles of Oxygen.
Using Avogadro's number to determine the number of oxygen atoms:
6.02 x 10 to the 23rd power atoms/mole multiplied by 2.17 moles = 13.06 X 10 to the 23rd or 1.31 X 10 to the 24th power of oxygen atoms.