What is the empirical formula of a compound that contains 29% Na, 41% S, and 30% O by mass?
Divide the percentages by the relative atomic mass (RAM) to get the relative number of atoms. Multiply by appropriate numbers to get whole numbers of atoms.
Na : 29/23 = 1.261
S : 41/32 = 1.281
O : 30/16 = 1.875
Now divide by 1.27 (average of 1.26 and 1.28)
Na : 1
S : 1
O : 1.48
Multiply by 2 to get whole numbers
Na : 2
S : 2
O 3
So the formula is
Na2S2O3
The chemical name is sodium thiosulphate. It is used in the titration with iodine, fixation of photographs in the 'wet' darkroom, and many other uses.
Well, let's break it down! We've got 29% Na, 41% S, and 30% O. Now, let me put on my clown hat and juggle some calculations!
Assuming we have a 100g sample of this compound, we have 29g of Na, 41g of S, and 30g of O.
Now let's find the moles of each element. The molar mass of Na is about 23 g/mol, S is about 32 g/mol, and O is about 16 g/mol.
So, the moles of Na is 29g / 23 g/mol = 1.26 mol, moles of S is 41g / 32 g/mol = 1.28 mol, and moles of O is 30g / 16 g/mol = 1.88 mol.
Now we need to divide each of these by the smallest number of moles, which is 1.26 mol.
So, we have Na: 1.26 mol / 1.26 mol = 1, S: 1.28 mol / 1.26 mol ≈ 1, and O: 1.88 mol / 1.26 mol ≈ 1.5.
But, we like whole numbers when it comes to empirical formulas, right? So, let's multiply each element by 2 to get whole numbers.
That gives us Na₂, S₂, and O₃.
Therefore, the empirical formula of the compound is Na₂S₂O₃. Voila!
To determine the empirical formula of a compound, you need to find the smallest whole number ratio of atoms present in the compound.
1. Start by assuming you have 100 grams of the compound. This means you have 29 grams of Na, 41 grams of S, and 30 grams of O.
2. Convert the mass of each element into moles. To do this, divide the mass by the molar mass of each element:
- Na: 29 g / 22.99 g/mol = 1.26 mol
- S: 41 g / 32.07 g/mol = 1.28 mol
- O: 30 g / 16.00 g/mol = 1.88 mol
3. Divide each mole value by the smallest mole value to find the mole ratio:
- Na: 1.26 mol ÷ 1.26 mol = 1
- S: 1.28 mol ÷ 1.26 mol ≈ 1.02
- O: 1.88 mol ÷ 1.26 mol ≈ 1.49
4. Multiply each atom's mole ratio by the smallest mole ratio to get whole number ratios. In this case, we can multiply by 100 to make them whole numbers:
- Na: 1 × 100 = 100
- S: 1.02 × 100 ≈ 102
- O: 1.49 × 100 ≈ 149
5. Write the empirical formula using the whole number ratios as the subscripts of each element:
The empirical formula of the compound is Na100S102O149. However, we can simplify it by dividing all the subscripts by the smallest subscript. In this case, dividing by 100 gives us the final empirical formula:
The simplified empirical formula is NaS1.02O1.49.
Therefore, the empirical formula of the compound is NaS1.02O1.49.
To find the empirical formula of a compound, you need to determine the simplest ratio of atoms present in the compound. This can be done by converting the given percentages into moles.
Here's how you can calculate the empirical formula for the given compound that contains 29% Na, 41% S, and 30% O by mass:
Step 1: Assume you have 100 grams of the compound. This means you have 29 grams of Na, 41 grams of S, and 30 grams of O.
Step 2: Convert the masses of each element to moles. To do this, divide the mass of each element by its molar mass.
The molar masses of Na, S, and O are:
Na: 22.99 g/mol
S: 32.07 g/mol
O: 16.00 g/mol
Converting the masses to moles:
Na: 29 g / 22.99 g/mol = 1.26 mol
S: 41 g / 32.07 g/mol = 1.28 mol
O: 30 g / 16.00 g/mol = 1.88 mol
Step 3: Divide each mole value by the smallest mole value to get the simplest, whole number ratio.
Dividing by 1.26 (the smallest mole value):
Na: 1.26 mol / 1.26 mol = 1
S: 1.28 mol / 1.26 mol ≈ 1
O: 1.88 mol / 1.26 mol ≈ 1.5
Step 4: Multiply each ratio by a whole number to get whole numbers.
Na: 1 x 2 = 2
S: 1 x 2 = 2
O: 1.5 x 2 = 3
Step 5: Write the empirical formula using the whole numbers obtained.
The empirical formula of the compound is Na2S2O3.