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Sergio039 [100]
2 years ago
14

An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical

and the molecular formula of the oxide'
Chemistry
2 answers:
sladkih [1.3K]2 years ago
7 0
Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help
ss7ja [257]2 years ago
3 0

<u>Answer:</u> The empirical and molecular formula for the given compound is P_2O_3 and P_4O_6 respectively.

<u>Explanation:</u>

We are given:

Percentage of P = 56.4 %

Percentage of O = 43.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 56.4 g

Mass of O = 43.6 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{56.4g}{31g/mole}=1.82moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{43.6g}{16g/mole}=2.725moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.82 moles.

For Phosphorus = \frac{1.82}{1.82}=1

For Oxygen = \frac{2.725}{1.82}=1.5

Converting the calculated moles into lowest whole number, we multiply the number of moles by '2'

Moles of Phosphorus = (1 × 2) = 2

Moles of oxygen = (1.5 × 2) = 3

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 3

Hence, the empirical formula for the given compound is P_2O_3

  • For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 220 g/mol

Mass of empirical formula = 110 g/mol

Putting values in above equation, we get:

n=\frac{220g/mol}{110g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

P_{(2\times 2)}O_{(3\times 2)}=P_4O_6

Hence, the empirical and molecular formula for the given compound is P_2O_3 and P_4O_6 respectively.

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How much potassium chloride is needed to make 0.500 m solution with 1.50 L of water?
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Answer:

55.9 g KCl.

Explanation:

Hello there!

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m=\frac{mol}{kilograms}

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4 0
3 years ago
Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Drag the appropriate items to their respective b
Sonja [21]

Answer:

Ag+(aq)+Br−(aq)→AgBr(s)                                NEGATIVE

CaCO3(s)→CaO(s)+CO2(g)2                           POSITIVE

NH3(g)→N2(g)+3H2(g)                                    POSITIVE

2Na(s)+Cl2(g)→2NaCl(s)                                 NEGATIVE

C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)           POSITIVE

I2(s)→I2(g)                                                        POSITIVE

Explanation:

We have to remember, to solve this problem, that the entropy of a gas is higher than that of a liquid which in turn  is higher than the solid. Therefore, comparing the reactants and products look for changes in the state of reactants and products. We also have to look for the increase or decrease of moles of each state based on the balanced chemical reaction.

Ag+(aq)+Br−(aq)→AgBr(s)

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Here we have a solid reactant and we have a solid product plus a gas product. The change in entropy is positive.

NH3(g)→N2(g)+3H2(g)

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2Na(s)+Cl2(g)→2NaCl(s)

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C3H8(g)+5O2(g)→3CO2(g) +4H2O(g)

The products are 7 mol of gas versus 6 mol of gas reactants and therefore entropy has increased.

I2(s)→I2(g)

1 mol solid I₂ goes into 1 mol gas making the change in  the entropy higher.

4 0
3 years ago
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