What is the empirical formula of a compound that contains 29% Na, 41% S, and 30% O by mass? ..My answer is NaSO and I am not sure... can anyone help me for it?????????
I'm relatively certain that it isn't NaSO.
Take 100 g sample which will give you
29 g Na
41 g S
30 g O
Now convert grams to moles. Remember moles = gram/molar mass.
moles Na = 29/23 =
moles S = 41/32 =
moles O = 30/16 =
You may want to look up the atomic masses of the elements Na, S, and O. I'm just using from my memory.
Now convert moles to the ratio of small whole numbers. The easy way to do that is to divide the smallest number by itself. That gives you 1.00 for it. Then divide all of the other numbers by the same small number. Round to whole numbers and that should be the subscripts that go with the formula. A word of caution: Don't round 0.5 or 0.3 or 0.25 to a whole number since the values usually are more accurate than that. That is, if you get 1.00 to 1.25, then the ratio would be 4:5 and not 1:1. Post your work if you get stuck.
To determine the empirical formula of a compound based on the given percentages, we need to convert the mass percentages into moles.
Let's assume we have 100 grams of the compound. Therefore, we have 29 grams of Na, 41 grams of S, and 30 grams of O.
The molar mass of Na is 22.99 g/mol, S is 32.06 g/mol, and O is 16.00 g/mol.
Next, we need to convert the mass of each element to moles by dividing the mass by its respective molar mass:
Na: 29 g / 22.99 g/mol ≈ 1.26 mol
S: 41 g / 32.06 g/mol ≈ 1.28 mol
O: 30 g / 16.00 g/mol ≈ 1.88 mol
Now we need to find the simplest whole number ratio by dividing each mole value by the smallest value (in this case, 1.26 mol):
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
Since all three elements have approximately the same ratio as 1:1:1.5, we can multiply each ratio by 2 to get whole numbers:
Na: 1 × 2 = 2
S: 1 × 2 = 2
O: 1.5 × 2 = 3
Therefore, the empirical formula of the compound is Na2SO3.
To determine the empirical formula of a compound given the percentages of its elements by mass, we need to follow a set of steps. Let's go through them to find the empirical formula of the compound:
Step 1: Assume we have a 100g sample of the compound. This means that within that sample, we would have 29g of Na, 41g of S, and 30g of O. We use grams because the percentages given are based on mass.
Step 2: Convert the given masses to moles. This can be done by dividing the mass of each element by its molar mass. The molar masses of the elements are Na: 22.99 g/mol, S: 32.06 g/mol, and O: 16.00 g/mol.
The moles of Na can be calculated as follows: 29g / 22.99 g/mol = 1.26 mol Na
The moles of S can be calculated as follows: 41g / 32.06 g/mol = 1.28 mol S
The moles of O can be calculated as follows: 30g / 16.00 g/mol = 1.88 mol O
Step 3: Determine the simplest whole-number ratio of the moles. To do this, divide each mole value by the smallest number of moles. In this case, the smallest number of moles is 1.26 mol Na.
Dividing all the moles by 1.26 gives us:
Na: 1.26 mol Na / 1.26 mol Na = 1 mol Na
S: 1.28 mol S / 1.26 mol Na ≈ 1.01 mol S
O: 1.88 mol O / 1.26 mol Na ≈ 1.49 mol O
Step 4: Write the empirical formula using the moles obtained. The empirical formula represents the simplest whole-number ratio of the elements in the compound.
Based on the ratio calculated above, the empirical formula would be NaSO. It should be noted that the ratio is not exact due to rounding during the calculations, so the empirical formula should be written as NaSO with roughly equal amounts of Na and S, and about 1.5 times the amount of O compared to Na.
Therefore, your answer of NaSO is correct!