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3 years ago

what is the empirical formula of a compound that contains 15.77% aluminum, 28.11% sulfur and 56.12% oxygen

Chemistry

1 answer:

Marizza181 [45]3 years ago

3 0

<h3>Answer:</h3>

Al₂(SO₄)₃

<h3>Explanation:</h3>

We are given percentage composition of elements in a compound;

Aluminium is 15.77%
Sulfur is 28.11 %
Oxygen is 56.12%

We are required to calculate the empirical formula of the compound.

Assuming the mass of the compound is 100 g then the masses of the elements is;

Aluminium = 15.77 g

Sulfur = 28.11

Oxygen = 56.12

We can determine the number of moles of each;

Moles of Aluminium = 15.77 g ÷ 26.98 g/mol

= 0.585 moles

Moles of sulfur = 28.11 g ÷ 32.07 g/mol

= 0.877 moles

Moles of Oxygen = 56.12 g ÷ 16.0 g/mol

= 3.5075 moles

But, the empirical formula is the simplest whole number ratio of elements in a compound.
Therefore; we need to get the ratio of moles of the above elements;

Aluminium : Sulfur : Oxygen

0.585 mol : 0.877 mol : 3.5075 mol

0.585/0.585 : 0.877/0.585 : 3.5075/0.585

1 : 1.5 : 6

But, we need whole number ratios, therefore;

= (1 : 1.5 : 6 ) × 2

= 2 : 3 : 12

Therefore; the formula of the compound is Al₂S₃O₁₂

The compound is written as Al₂(SO₄)₃

Thus, the empirical formula of the compound is Al₂(SO₄)₃

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The name iron-60 also tells us that this particlar isotope's mass number is 60.
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3 years ago

The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to

tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of $CuFeS_2$ = 183.511 g/mole

First we have to calculate the moles of Cu.

$\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles$

The moles of Cu = 4.7209 moles

From the given chemical formula, $CuFeS_2$ we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of $CuFeS_2$ = 4.4209 moles

Now we have to calculate the mass of $CuFeS_2$ .

Mass of $CuFeS_2$ = Moles of $CuFeS_2$ × Molar mass of $CuFeS_2$ = 4.4209 moles × 183.511 g/mole = 866.337 g

Mass of $CuFeS_2$ = 866.337 g = 0.8663 Kg (1 Kg = 1000 g)

Therefore, 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

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3 years ago

In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the

densk [106]

Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

Q sln = mCΔT

∴ m sln = m Ba(OH)2 + m HCl

∴ molar mass Ba(OH)2 = 171.34 g/mol

∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol

⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g

∴ molar mass HCl = 36.46 g/mol

∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol

⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g

⇒ m sln = 3.084 g + 1.313 g = 4.3966 g

specific heat (C):

∴ C sln = C H2O = 4.18 J/g°C

∴ ΔT = 26.83°C - 22.74°C = 4.09°C

heat absorbed (Q):

⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)

⇒ Q sln = 75.165 J

8 0

3 years ago

How are elements arranged in the Periodic Table? A. by their properties B. alphabetically C. by how much they are worth D. in th

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Im on the same test.

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3 years ago

Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1

timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

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Step 2: [Ni(H₂O)₄(en)]²⁺ + 1en → [Ni(H₂O)₂(en)₂]²⁺ ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en → [Ni(en)₃]²⁺ ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺ + 3en → [Ni(en)₃]²⁺ ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ = - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0

3 years ago