Chemistry, 21.07.2019 03:31 Qpaoswp34

When a 30.98-g sample of phosphorus reacts with oxygen, a 71.00-g sample of phosphorus oxide is formed. what is the percent composition of the compound? what is the empirical formula for this compound?

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Answer from: jourdyno

- P = 43.63% and O = 56.37% and the empirical formula is P_2O_5

Solution:- mass of Phosphorous = 30.98 g

mass of product(phosphorous oxide) = 71.00 g

mass of oxygen = 71.00 - 30.98 = 40.02 g

percentage of phosphorous in the compound = (30.98/71.00)100 = 43.63%

percentage of oxygen in the compound = (40.02/71.00)100 = 56.37%

moles of P = 30.98 g x ( 1mol/30.97 g) = 1.00 mol

moles of O = 40.02 g x (1 mol/ 16.00 g) = 2.50 mol

Ratio of moles of P to O is 1.00:2.50

The whole number ratio is 2:5

So, the empirical formula of the phosphorous oxide formed is P_2O_5

Answer from: Monicaamm12983

Answer : The percent composition of P = 43.63 % and O = 53.37 % .

Empirical formula of Phosphorous oxide = P₂O₅

Part A : Percent composition :

It is percent of each element present in compound . It is given by formula :

$Percent composition = \frac{mass of element}{ total mass of compound } * 100$

Given : Mass of Phosphorous (P ) = 30.98 g

Mass of Compound Phosphorous oxide = 71.00 g

Mass of Oxygen (O) = mass of compound - mass of P

= 71.00 g - 30.98 g = 40.02 g

$Percent composition of P = \frac{mass of P }{ mass of compound}$ $= \frac{30.98 g }{71.00 g} * 100$

Percent composition of P = 43.63 %

$Percent composition of O = \frac{mass of Al}{Mass of compound } * 100 = \frac{40.02 g }{71.00 g} * 100$

Percent composition of O = 53.37 %

Part B : Empirical formula of Phosphorous Oxide.

Empirical formula is formula which shows the proportion of element present in a compound . Following are the steps to calculate empirical formula of an compound :

Step 1 : Find masses of each element .

Mass of P = 30.98 g

Mass of O = 40.02 g

Step 2 : Conversion of masses of element to its mole .

$Mole = \frac{mass }{molar mass }$

Given : Mass of P = 30.98 g $Molar mass of P = 30.97 \frac{g}{ mol}$

$Mole of P = \frac{30.98 g }{30.97\frac{g}{mol} } = 1 mol$

Given: Mass of O = 40.02 g $Molar mass of O = 15.99\frac{g}{mol}$

$Mole of O = \frac{40.02 g}{15.99\frac{g}{mol}} = 2.5 mol$

Step 3 : Finding Ratio of mole .

In this step ratio is found by dividing each mole by smallest mole .Since mole of P is smaller , so this will be used for division.

$Ratio of mole of P= \frac{mole of P }{mole of P } = \frac{1 mol}{1 mol} = 1$

$Ratio of mole of O = \frac{ mole of O }{mole of P } = \frac{2.5 }{1 } = 2.5$

Hence , Ratio of P : O = 1 : 2.5

Since the ratio is in fraction , it need to be converted to whole number . So we multiply the ratio by such a minimum number which gives us a whole number ratio. This step is skipped if the ratio already comes in whole number.

On multiplication the ratio by 2 :

Ratio of P : O = ( 1 : 2.5 ) * 2 = 1 * 2 : 2.5 * 2

Ratio of P : O = 2 : 5

Step 4 : Writing the empirical formula

The ratio of P and O is 2: 5 , which gives the empirical formula of compound as P₂O₅ .

Answer from: Quest

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

Answer from: Quest

Heat is involved in chemical change in two ways; chemical changes either result in heat being added to or removed from an object, and chemical changes require a certain amount of heat before they can take place.

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When a 30.98-g sample of phosphorus reacts with oxygen, a 71.00-g sample of phosphorus oxide is form...