when a student removed the crucible cover afer the first strong heating in creating magnesium oxide, some of the white was knocked from the cover to the floor and was lost. did this loss cause the calculated number of moles of oxygen in the compound to be too high or too low?
Too low but you can figure that yourself.
Let's say we start with 5 g Mg and prepared MgO properly. We will get the following numbers (I calculated what they should be).
mass Mg = 5.00 g
mass MgO = 8.29 g
mass Oxygen = 8.29- 5.00 = 3.29
So:
moles Mg = 5.00/24.3 = 0.206
moles oxygen = 3.29/16 = 0.206
and the ratio is 1:1 for the formula of MgO.
Now that we have made up what SHOULD happen, we can calculate what happens if we lose some of the white powder.
mass Mg = 5.00 g
mass oxide = 6.65 g (instead of 8.29)
mass oxygen = 6.65 - 5.00 = 1.65 g
Now we determine moles
moles Mg = 5.00/24.3 = 0.206
moles oxygen = 1.65/16 = 0.103
So the formula would be Mg2O.
Thus losing some of the white powder would make moles oxygen too low.
Well, if some of the white stuff was lost, then I guess you could say the number of moles of oxygen just went up in smoke! Okay, bad joke. But seriously, if some of the white substance was knocked off and lost, it means there was less of it left to be analyzed. So, the calculated number of moles of oxygen in the compound would be too low. That's one way to make a chemistry experiment a bit more exciting, right?
If some of the white substance (presumably magnesium oxide) was knocked from the crucible cover and lost, this loss would cause the calculated number of moles of oxygen in the compound to be too low. This is because some of the magnesium oxide, which contains oxygen, was lost along with the white substance.
To determine whether the loss of white substance from the crucible cover caused the calculated number of moles of oxygen in the compound to be too high or too low, we need to understand the stoichiometry of the reaction.
In the creation of magnesium oxide (MgO), the balanced chemical equation is:
2 Mg(s) + O2(g) -> 2 MgO(s)
From the equation, we can see that for every 2 moles of magnesium (Mg), 2 moles of magnesium oxide (MgO) are formed. Likewise, the stoichiometry tells us that 1 mole of oxygen gas (O2) is required to react with 2 moles of magnesium to form 2 moles of magnesium oxide.
Now, let's consider the situation where some of the white substance (MgO) was lost from the crucible cover. If the lost white substance happened to be magnesium oxide (MgO), then the overall number of moles of oxygen would be lower than expected. This is because the lost MgO contains oxygen.
So, in this case, the loss of white substance would cause the calculated number of moles of oxygen in the compound to be too low.