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2 years ago

The number of grams per mole of something is also known as_______

Chemistry

2 answers:

DaniilM [7]2 years ago

7 0

<h2>Answer:</h2>

<u>Molar mass</u> is also known as number of grams/mole of something.

<h2>Explanation:</h2>

A substance's molar mass is defined as substance mass divided by the substance amount in moles. It is expressed in Kg/mol but gram/mol is used for the convenience for study.

It is often used in chemistry for the conversion of mass in grams to substance moles. The relation that is used to convert the gram top mole is that the molar mass is equal to the molecular mass in grams.

Sergeeva-Olga [200]2 years ago

5 0

Answer:

Molar mass

Explanation:

This is a counting unit which represents the mass in grams of a substance that make up one mole of the substance. This mass is calculated as follows:

Molar mass = Mass/ Number of moles

Units: grams/mol

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3 years ago

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A 0.288 g sample of an unknown monoprotic acid is dissolved in water and titrated with a 0.115 M NaOH solution. After the additi

Sholpan [36]

Answer:

74.0 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA + NaOH ⇒ NaA + H₂O

Step 2: Calculate the reacting moles of NaOH

At the equivalence point, 33.83 mL of 0.115 M NaOH react.

0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol

Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.

Step 4: Calculate the molar mass of the acid

3.89 × 10⁻³ moles of HA have a mass of 0.288 g.

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

<u>Answer:</u> The empirical formula for the given compound is $C_3H_6O$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

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

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

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

<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 $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

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

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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3 years ago