Chemistry, 03.07.2019 20:00 ratpizza

What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?

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Answer from: jessiegarrison1

Given:

% Al = 35.94

% S = 64.06

To determine:

Empirical formula of a compound with the above composition

Explanation:

Atomic wt of Al = 27 g/mol

Atomic wt of S = 32 g/mol

Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g

# moles of Al = 35.94/27 = 1.331

# moles of S = 64.06/32 = 2.002

Divide by the smallest # moles:

Al = 1.331/1.331 = 1

S = 2.002/1,331 = 1.5 ≅ 2

Empirical formula = AlS₂

Answer from: sarahidan

The empirical formula for the given compound is Al_2S_3

Explanation:

Let us assume that the mass of the compound be 100 grams.

So, the percentage of the elements will be equal to their respective masses.

Mass of aluminium = 35.94 g

Mass of sulfur = 64.06 g

To find the empirical formula for a compound, we follow some steps:

Step 1: Converting the given masses of the elements into moles.

To calculate moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of aluminium = $\frac{35.94g}{27g/mol}=1.33mol$

Moles of sulfur = $\frac{64.06g}{32g/mol}=2mol$

Step 2: Now, dividing each calculated moles by the smallest number of moles, which is 1.33 moles, to calculate the mole ratio of elements.

Mole ratio of aluminium = $\frac{1.33}{1.33}=1$

Mole ratio of sulfur = $\frac{2}{1.33}=1.5$

Step 3: Converting the mole ratio into whole number ratio.

Multiplying both the ratios by 2, to make both the mole ratios to be a whole number.

Mole ratio of aluminium = $1\times 2=2$

Mole ratio of sulfur = $1.5\times 2=3$

Step 4: Writing each mole ratio as the subscripts of the elements to make the empirical formula.

Empirical formula for the given compound is Al_2S_3

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

Electrons are transferred from atoms of to atoms of this transfer makes the sodium atoms and the

Answer from: Quest

4d^55s^1 ( third choice)

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