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vladimir2022 [97]
3 years ago
12

2) A sample of table sugar (sucrose C12H22O11) has a mass of 1.202g

Chemistry
1 answer:
Basile [38]3 years ago
6 0

Answer:

A) 0.003512 moles

B)

Carbon =  0.04214 moles

Hydrogen = 0.07726 moles

Oxygen = 0.03863 moles

C) Carbon = 2.54*10^22 atoms

Hydrogen =  4.65 *10^22 atoms

Oxygen = 2.33 * 10^22 atoms

Explanation:

Step1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = mass / molar mass

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate the moles of each element in C12H22O11.

For 1 mol of sucrose we have 12 mol of carbon, 22 mol of hydrogen and 11 mol of oxygen

Carbon = 12*0.003512 = 0.04214 moles

Hydrogen = 22* 0.003512 = 0.07726 moles

Oxygen = 11* 0.003512 = 0.03863 moles

Step 4: Calculate the number of atoms of each type in C12H22O11.

Carbon = 0.04214 moles * 6.022*10^23/mol = 2.54*10^22 atoms

Hydrogen = 0.07726 moles * 6.022*10^23/mol = 4.65 *10^22 atoms

Oxygen = 0.03863 moles * 6.022*10^23/mol = 2.33 * 10^22 atoms

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Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic.
Dmitrij [34]

Answer :

(a) The molar mass of C_2H_3Cl_3O_2 is, 165.5 g/mole

(b) The moles of C_2H_3Cl_3O_2 is, 3.02 moles

(c) The mass in grams of 2.0\times 10^{-2} mole chloral hydrate is, 3.31 g

(d) The number of chlorine atoms in 5.0 g chloral hydrate is, 5.4\times 10^{22}

(e) The mass of chloral hydrate will be, 1.55 g

(f) The mass of exactly 500 molecules of chloral hydrate is, 1.99\times 10^{23}

Explanation :

(a) To calculate the molar mass of chloral hydrate.

The formula of chloral hydrate is, C_2H_3Cl_3O_2

Atomic mass of carbon = 12 g/mole

Atomic mass of hydrogen = 1 g/mole

Atomic mass of oxygen = 16 g/mole

Atomic mass of chlorine = 35.5 g/mole

Now we have to determine the molar mass of chloral hydrate.

\text{Molar mass of }C_2H_3Cl_3O_2=2(12g/mole)+3(1g/mole)+3(35.5g/mole)+2(16g/mole)=165.5g/mole[/tex]

The molar mass of C_2H_3Cl_3O_2 is, 165.5 g/mole

(b) Now we have to determine the moles of C_2H_3Cl_3O_2.

\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{500.0g}{165.5g/mole}=3.02moles

The moles of C_2H_3Cl_3O_2 is, 3.02 moles

(c) Now we have to determine the mass in grams of 2.0\times 10^{-2} mole chloral hydrate.

\text{Mass of }C_2H_3Cl_3O_2=\text{Moles of }C_2H_3Cl_3O_2\times \text{Molar mass of }C_2H_3Cl_3O_2

\text{Mass of }C_2H_3Cl_3O_2=(2.0\times 10^{-2}mole)\times (165.5g/mole)=3.31g

The mass in grams of 2.0\times 10^{-2} mole chloral hydrate is, 3.31 g

(d) To calculate the number of chlorine atoms are in 5.0 g chloral hydrate.

First we have to determine the moles of C_2H_3Cl_3O_2.

\text{Moles of }C_2H_3Cl_3O_2=\frac{\text{Mass of }C_2H_3Cl_3O_2}{\text{Molar mass of }C_2H_3Cl_3O_2}=\frac{5g}{165.5g/mole}=0.03moles

Now we have to calculate the number of chlorine atoms in chloral hydrate.

In C_2H_3Cl_3O_2, there are, 2 carbon atoms, 3 hydrogen atoms, 3 chlorine atoms and 2 oxygen atoms.

As, 1 mole of C_2H_3Cl_3O_2 contains 3\times 6.022\times 10^{23} chlorine atoms

So, 0.03 mole of C_2H_3Cl_3O_2 contains 0.03\times 3\times 6.022\times 10^{23}=5.4\times 10^{22} chlorine atoms

The number of chlorine atoms in 5.0 g chloral hydrate is, 5.4\times 10^{22}

(e) To calculate the mass of chloral hydrate would contain 1.0 g Cl.

As, 3\times 35.5g of chlorine present in 165.5 g of C_2H_3Cl_3O_2

So, 1 g of chlorine present in \frac{165.5}{3\times 35.5}=1.55g of C_2H_3Cl_3O_2

The mass of chloral hydrate will be, 1.55 g

(f) To calculate the mass of exactly 500 molecules of chloral hydrate.

As, 6.022\times 10^{23} molecules of chloral hydrate has 165.5 g mass of chloral hydrate

So, 500 molecules of chloral hydrate has \frac{6.022\times 10^{23}}{500}\times 165.5=1.99\times 10^{23} mass of chloral hydrate

The mass of exactly 500 molecules of chloral hydrate is, 1.99\times 10^{23}

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4 years ago
What would changing the number of electrons in a atom change
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Answer:

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