Design an illustrative graphic depicting a chemistry concept related to empirical formulas. For instance, envision some empty beakers and analytical balances ready for compound analysis on a laboratory table. Zoom in on one beaker containing a crystal-like substance, indicating a compound under examination. Also, visualize a decomposing alkaline battery with copper terminals and manganese dioxide within, a clear glass vial holding a volatile substance representing a compound with chlorine and oxygen, and a finely powdered reddish-brown substance in a petri dish representing magnetic iron oxide.

Determine the empirical formula for each substance.

1. A dead alkaline battery is found to contain a compund of Mn and O. Its analysis gives 69.6% Mn and 30.4% O.

2. A compound is 38.77% Cl and 61.23 %O.

3. Magnetic iron oxide is 72.4 % iron and 27.6% oxygen.

Assume 100 grams. I will do one, what about 3 as an example

YOu have 72.4g Fe, 27.6gO
this represents in moles...

Fe=72.4/55.8= 1.30 moles
O= 27.6/15.99=1.73 moles

Now divide each by the lowest moles (we are looking for whole number ratios)

Fe=1.30/1.30=1
O=1.73/1.30=1.5
But we cant write a formula as
FeO1.5
so we multiply the numbers by an integer until all the subscripts are whole numbers, in this case multiply by 2 to get
Fe2O3

Bob Pursley worked one for you and I worked another one. You can follow our examples and do the others. You need to learn to do these yourself.

All of these are done the same way.

Take 100 g sample. For Mn and O, that will give you 69.6 g Mn and 30.4 g oxygen.
Now convert to mols.
moles Mn = 69.4/atomic mass Mn.
moles oxygen = 30.4/atomic mass O.
Now you want to find the formula. Remember that Dalton suggested that molecules are made of atoms in the ratio of small whole numbers. Therefore, you want to convert the moles you have above into small whole number. The easy way to do that is to divide all of the moles by the smallest number. That will give you 1.000 for the smallest number and the other should be very close to a whole number . Round to a whole number unless it is something like 1 to 1.5 in which case multi;ly both by 2 to obtain small whloe numbers. Post your work if you get stuck.

Oh, I love playing with chemical formulas! Let's have some fun figuring these out:

1. Since the percentages given are for Mn and O, we can assume that there's 69.6g of Mn and 30.4g of O in the compound. The next step is to convert these grams into moles. The molar mass of Mn is approximately 54.94 g/mol, and that of O is roughly 16.00 g/mol.

Calculating the moles:
Mn: 69.6 g / 54.94 g/mol ≈ 1.267 mol
O: 30.4 g / 16.00 g/mol ≈ 1.900 mol

Now we can simplify by dividing both of these values by the smaller one (1.267 mol).

Mn: 1.267 mol / 1.267 mol = 1
O: 1.900 mol / 1.267 mol ≈ 1.500

So, the empirical formula for this compound is MnO₁.₅ (approximately).

2. Similar to the previous question, we'll start by converting the percentages into grams. Assuming we have 38.77g of Cl and 61.23g of O, let's calculate the moles.

Molar mass of Cl is about 35.45 g/mol, and that of O is still around 16.00 g/mol.

Moles:
Cl: 38.77 g / 35.45 g/mol ≈ 1.094 mol
O: 61.23 g / 16.00 g/mol ≈ 3.827 mol

Dividing both by the smaller value (1.094 mol), we get:
Cl: 1.094 mol / 1.094 mol = 1
O: 3.827 mol / 1.094 mol ≈ 3.5

So, the empirical formula for this compound is ClO₃.₅ (approximately).

3. Lastly, let's determine the empirical formula for magnetic iron oxide.
Given that it consists of 72.4% iron and 27.6% oxygen, we can assume 72.4g of iron and 27.6g of oxygen.

Molar mass of iron is about 55.85 g/mol, and oxygen is about 16.00 g/mol.

Moles:
Iron: 72.4 g / 55.85 g/mol ≈ 1.299 mol
Oxygen: 27.6 g / 16.00 g/mol ≈ 1.725 mol

Dividing both by the smaller value (1.299 mol), we obtain:
Iron: 1.299 mol / 1.299 mol = 1
Oxygen: 1.725 mol / 1.299 mol ≈ 1.33

Therefore, the empirical formula for magnetic iron oxide is FeO₁.₃₃ (approximately).

Remember, empirical formulas show the simplest ratio of elements in a compound. Enjoy the wonderful world of chemistry, my friend!

To determine the empirical formula for each substance, we need to find the ratio of the elements present in the compound. Here's how to do it for each case:

1. A dead alkaline battery is found to contain a compound of Mn and O. Its analysis gives 69.6% Mn and 30.4% O.

First, assume we have 100 grams of the compound. So we have 69.6 grams of Mn and 30.4 grams of O.

Next, we need to convert the mass of each element to the number of moles using their molar masses. The molar mass of Mn is 54.938 g/mol, and the molar mass of O is 15.999 g/mol.

For Mn:
Number of moles of Mn = Mass of Mn / Molar mass of Mn
= 69.6 g / 54.938 g/mol
≈ 1.266 moles

For O:
Number of moles of O = Mass of O / Molar mass of O
= 30.4 g / 15.999 g/mol
≈ 1.9 moles

Now, we need to find the ratio of moles between Mn and O by dividing both moles by the smallest mole value:
Mn:O ≈ 1.266 moles / 1.266 moles : 1.9 moles / 1.266 moles
≈ 1 : 1.5

Therefore, the empirical formula for the compound in the dead alkaline battery is MnO1.5.

2. A compound is 38.77% Cl and 61.23% O.

Assuming we have 100 grams of the compound, we have 38.77 grams of Cl and 61.23 grams of O.

Next, we convert the mass of each element to the number of moles. The molar mass of Cl is 35.453 g/mol, and the molar mass of O is 15.999 g/mol.

For Cl:
Number of moles of Cl = Mass of Cl / Molar mass of Cl
= 38.77 g / 35.453 g/mol
≈ 1.093 moles

For O:
Number of moles of O = Mass of O / Molar mass of O
= 61.23 g / 15.999 g/mol
≈ 3.826 moles

To find the ratio of moles between Cl and O, we divide both moles by the smallest mole value:
Cl:O ≈ 1.093 moles / 1.093 moles : 3.826 moles / 1.093 moles
≈ 1 : 3.5

Therefore, the empirical formula for the compound is ClO3.5.

3. Magnetic iron oxide is 72.4% iron and 27.6% oxygen.

Assuming we have 100 grams of the compound, we have 72.4 grams of iron and 27.6 grams of oxygen.

Next, we convert the mass of each element to the number of moles. The molar mass of iron is 55.845 g/mol, and the molar mass of oxygen is 15.999 g/mol.

For iron:
Number of moles of iron = Mass of iron / Molar mass of iron
= 72.4 g / 55.845 g/mol
≈ 1.297 moles

For oxygen:
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
= 27.6 g / 15.999 g/mol
≈ 1.725 moles

To find the ratio of moles between iron and oxygen, we divide both moles by the smallest mole value:
iron:oxygen ≈ 1.297 moles / 1.297 moles : 1.725 moles / 1.297 moles
≈ 1 : 1.333

Therefore, the empirical formula for the compound is FeO1.333, which can be simplified to FeO4/3.

can you help with the others as well?