Asked by Betsy
A 2.000 g sample of magnesium was burned in air to form an oxide.
After the product was purified it was found to have a mass of 3.317 g. What is the empirical formula for the product?
After the product was purified it was found to have a mass of 3.317 g. What is the empirical formula for the product?
Answers
Answered by
Betsy
This is how I tried to answer...
1 mol of Mg= 24.3g
1 mol of O= 16.0 g
2.0 g Mg = 0.0823 mol Mag
1.317 g O (3.317-2.0) = 0.0823 mol O
empirical formula is MgO
1 mol of Mg= 24.3g
1 mol of O= 16.0 g
2.0 g Mg = 0.0823 mol Mag
1.317 g O (3.317-2.0) = 0.0823 mol O
empirical formula is MgO
Answered by
Dr Russ
Looks OK to me. Just to be picky watch what you are doing with the units and balancing equations, for example
2.0 g Mg = 0.0823 mol Mg
does not balance in terms of number or units.
Similarly
1.317 g O (3.317-2.0) = 0.0823 mol O
does not balance in terms of number or units.
Better as
Number of moles of magnesium is
2.0 g / 24.3 g mol^-1= 0.0823 mol
and number of moles of oxygen is
1.317 g /16 g mol^-1 = 0.0823 mol
2.0 g Mg = 0.0823 mol Mg
does not balance in terms of number or units.
Similarly
1.317 g O (3.317-2.0) = 0.0823 mol O
does not balance in terms of number or units.
Better as
Number of moles of magnesium is
2.0 g / 24.3 g mol^-1= 0.0823 mol
and number of moles of oxygen is
1.317 g /16 g mol^-1 = 0.0823 mol
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