Focus 1 Complete the diagram shown in Figure 5.1 to work out...

Focus 1 Complete the diagram shown in Figure 5.1 to work out the formula mass of the iron oxide in the ore magnetite. (Ar: Fe = 56, O = 16.) Then use the following steps to work out the percentage by mass of iron in this ore. The formula of magnetite is Fe3O4 The atoms present are Fe ........ Fe O ........ O ........ The relative atomic mass is ........ + ........ + ........ + ........ + ........ + ........ + ........ = ........ Figure 5.1: Chemical formula of magnetite, a form of iron oxide. - The relative formula mass of the iron oxide (Fe3O4) = ................... - In this formula, there are ................... atoms of iron, Fe. - The relative mass of ................... Fe = ................... - This means that in ...................g of Fe3O4 there are ................... g of iron. - So 1 g of Fe3O4 contains ................... g of iron. - So 100 g of Fe3O4 contains ................... g of iron. - In other words, the percentage (%) of iron in Fe3O4 = ...................%. 2 Oxalic acid is an organic acid present in rhubarb and some other vegetables. The composition by mass of oxalic acid is given below, and it has an Mr of 90. carbon = 26.7% hydrogen = 2.2% oxygen = 71.1% a What is the empirical formula of oxalic acid? b Calculate the molecular formula of the acid .................................................................................................................................... 3 A volatile arsenic compound containing arsenic, carbon and hydrogen has the following composition by mass: arsenic = 62.5% carbon = 30.0% hydrogen = 7.5% Calculate the empirical formula of this compound.

Transcript

00:01 So for the first part of this question, we are trying to find the formula mass and eventually the percent of iron in this compound.

00:11 So for the first part of this compound, we are dividing it up to see how many irons and how many oxygens are in this compound.

00:21 So the subscript tells you how many of each atom are in this compound.

00:25 So that three subscript of iron means that we have three irons in the compound, and the four by the oxygen means that we have four oxygens.

00:38 Now when we are calculating the formula mass, we add up the weights of all of the atoms in that compound.

00:48 So for that long addition problem below, we are going to add up the molecular weight of each iron, which is 56 plus 56 plus 56.

01:05 For each of our three irons and then the molecular weight of oxygen is 16.

01:12 So we're going to add four 16s because we have four oxygens.

01:21 If you add that all together, you should get 232.

01:31 So our formula mass for the compound is 232 grams per mole.

01:41 Another way that you could write this right in an addition problem, since we have three irons, we could take three times the atomic weight of iron plus we have four oxygen.

01:55 So we would take four times the molecular weight oxygen and we will get the same answer.

02:03 For the next part of this question, we have to figure out how many atoms of iron are in this formula.

02:15 So we have, since we have three irons, that means we have three moles of iron in this compound.

02:29 So we have three moles of iron in the compound.

02:35 To convert that into atoms, we need to use avogadro's number.

02:39 So we take that times avogadro's number, which is 6 .022 times 10 to the 23rd.

02:45 And that is how many atoms there are per one mole.

02:51 The moles cancel out, leaving us with atoms of iron.

02:56 And you should get 1 .81 times 10 to the 24th atoms of iron.

03:12 Next, we have to figure out the relative mass of iron.

03:17 So i'm assuming that means that for the mass of three moles of iron or the total mass of iron in this compound.

03:27 So we have three moles of iron, right, which we get from that subscript.

03:31 Since there's three irons, we have three moles of iron because these numbers, these subscripts are just the ratio of moles between the two atoms in the compound.

03:42 So we have three moles of iron.

03:44 And if we take that times the atomic weight of iron, which they gave us as 56 grams of iron for every one mole of iron.

03:58 The mole is going to cancel.

04:00 So we take three times 56 and you should get 168 grams of iron.

04:13 So that is going to be the mass of iron.

04:15 So the mass of three moles of iron is equal to 168 grams of iron.

04:33 So this means that if we look at the formula mass, that in 232 grams of the iron oxide, we have 168 grams of iron.

04:58 Iron.

05:04 So that means for every one gram of the iron oxide, how many grams of iron do we have? so this can be set up as a simple ratio.

05:17 So if we have 168 grams of iron, for every 232 grams of the oxide, we can set it up as a ratio.

05:35 So we're trying to figure out how many grams of iron would be in 1 gram of the iron oxide.

05:44 We set up as a ratio because no matter how many grams of the iron oxide we have, we would have the direct ratio of grams of iron.

05:57 They keep the ratio no matter how much the mass changes because this formula always stays the same.

06:10 So if we do a little quick algebra, we can say right where we do the times the diagonals.

06:21 If you go way, way back to algebra, so 232 times x equals 168 times 1, which is 168.

06:30 So x is going to be 168 divided by 232, which would be 0 .724.

06:41 X is our grams of iron in one gram of iron oxide.

06:48 So if we go back for every one gram of iron oxide, we're going to have 0 .724 grams of iron.

07:02 So then if we go and we say we have 100 grams of iron oxide, the ratio is going to stay the same.

07:10 So if we take 1 times 100, we get 100.

07:13 So we're going to have to take 0 .724 times 100, which gets us 72.

07:19 0 .4 grams of iron, right? the ratio just stays the same.

07:28 So in other words, the percentage of iron.

07:30 So out of 100 grams, we have 72 .4 grams of iron.

07:34 So that means the percentage of iron and the iron oxide would be 72 .4%.

07:42 Now let's move on to the next question where we're trying to find the empirical formula and the molecular formula of oxen.

07:59 So we're given the percentages of each carbon, hydrogen, and oxygen, and then we're told that the formula mass is 90 grams per mole.

08:15 So remember that the equation for figuring out the moles of each atom in the formula or the percent mass.

08:30 We know that the percent composition of the atom is equal to n, which is the number of moles of the atom, times the atomic weight of atom divided by the formula mass of the compound.

08:51 So we can use this formula and work backwards to figure out how many of each element are in the final compound.

09:02 So they've already given us the formula mass, which is 90 grams per mole.

09:07 They've given us the percentage of each atom already.

09:12 N is what we're trying to find out, which is how many of that atom are in the compound.

09:17 And then the atomic weight, we can just get from the periodic table.

09:21 So from the periodic table, we know that carbon is 12 grams per mole...

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