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sergey [27]
3 years ago
13

A hydrocarbon contains C,H, and F atoms. A 5.43 g sample of this hydrocarbon was analyzed and the mass of 2.35 g of carbon and 0

.294 g of hydrogen. The molar mass of this compound was measured between 219-225 g/mol, what is the molecular formula of the hydrocarbon that contains C,H and F?
Chemistry
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

C8H12F6

Explanation:

To solve this question we need to find the moles of each atom in order to find the empirical formula (The empirical formula is defined as the simplest whole-number ratio of atoms present in a molecule). Using the empirical formula and the molar mass we can find molecular formula as follows:

<em>Moles C:</em>

2.35g * (1mol / 12g) = 0.1958 moles C

<em>Moles H:</em>

0.294g (1mol /1g) = 0.294 moles H

<em>Moles F -Molar mass: 19.0g/mol-: </em>

Mass F: 5.43g - 2.35g C - 0.294g H = 2.786g F * (1mol / 19.0g) = 0.1466 moles F

The moles of atoms dividing in the moles of F (Lower number of moles) produce the simplest ratio as follows:

C = 0.1958mol C / 0.1466mol F = 1.33

H =0.294mol H / 0.1466mol F = 2

F = 0.1466 mol F / 0.1466mol F = 1

As the empirical formula requires whole numbers, this ratio multiplied 3 times:

C = 4

H = 6

F=3

The empirical formula is:

C4H6F3

With molar mass of:

4C = 12*4 = 48

6H = 1*6 = 6

3F = 19*3 = 57

The molar mass is: 48g/mol + 6g/mol + 57g/mol = 111g/mol

As we know the molecule has a molar mass between 219-225g/mol and the empirical formula is 111g/mol, 2 times empirical formula will produce a molecule with the molar mass of the molecule, that is:

<h3>C8H12F6</h3>
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A sample of C3H8 has 6.72 x 10^24 H atoms. <br> What is the total mass of the sample?
gregori [183]
The answer is 492.8 g


1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample. 

1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole    

6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n

n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol


2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:

Ar(C) = 12 g/mol

Ar(H) = 1 g/mol

Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol



3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:

m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g

6 0
3 years ago
How many grams of HNO3 are produced when 59.0 g of NO2 completely reacts?
Anettt [7]

Answer:

53.7 grams of HNO3 will be produced

Explanation:

Step 1: Data given

Mass of NO2 = 59.0 grams

Molar mass NO2 = 46.0 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 59.0 grams / 46.0 g/mol

Moles NO2 = 1.28 moles

Step 4: Calculate moles HNO3

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3

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3 0
3 years ago
How do I solve this problem?
JulijaS [17]
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
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3 years ago
What is an alkaline earth metal that has 5 shells?
brilliants [131]

Answer:

The elements in the alkaline earth metals group; beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra), have two electrons in their outer electronic shell.

Explanation:

3 0
3 years ago
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If 30 ml of water containing 15 mg of salicylate is extracted with 30 ml of chloroform, 2.59 mg of salicylate remains in the wat
In-s [12.5K]

Answer:

Distribution coefficient: 4.79

Explanation:

Distribution coefficient is the ratio between equilibrium concentration of non-aqueous phase and aqueous phase where both solvents are inmiscible. The equation for the problem is:

Distribution coefficient: Concentration in chloroform / Concentration in Water

<em>Concentration in water: 2.59mg / 30mL = 0.08633mg/mL</em>

<em>Concentration in chloroform: (15mg-2.59mg) / 30mL = 0.4137mg/mL</em>

<em />

Distribution coefficient: 0.4137mg/mL / 0.08633mg/mL

<h3>Distribution coefficient: 4.79</h3>
8 0
3 years ago
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