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

im not exactly sure but i think its b!

Answer from: Quest

answer: any goods being imported from another country may be subject to issued by the federal government are tariffs

explanation:

tariffs are a tax on anything that are imported from abroad. government use them to protect and boost their local industries and the thought is that they encourage consumers to buy local instead of importing and they will hope will protect jobs and economy in the country that implements them

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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...

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