Which group has the greatest number of valence electrons?

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

Alexxandr [17]

Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: 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 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

4 0

3 years ago

5. A sample of pure acetyl salicylic acid had a mass of 0.268 grams. It was dissolved in NaOH and diluted to 500.00mL in a volum

Elza [17]

Answer:

a. 5.36x10⁻⁴ g/mL

b. 4.29x10⁻⁵ g/mL

Explanation:

As the units for concentration are not specified, I'll respond using g/mL.

a. We divide the sample mass by the final volume in order to calculate the concentration:

0.268 g / 500 mL = 5.36x10⁻⁴ g/mL

b. We can use C₁V₁=C₂V₂ for this question:

8.00 mL * 5.36x10⁻⁴ g/mL = C₂ * 100.00 mL

C₂ = 4.29x10⁻⁵ g/mL

4 0

3 years ago

2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)

Paha777 [63]

<h3>Answer:</h3>

16.7 g H₂O

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

Step 2: Identify Conversions

[RxN] 2 mol NaOH → 1 mol H₂O

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})$
Multiply/Divide: $\displaystyle 16.6685 \ g \ H_2O$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

16.6685 g H₂O ≈ 16.7 g H₂O

6 0

3 years ago

Are the particles moving faster or slower as time goes on? (Diagram F)

klasskru [66]

I YHINK ITTTSSS SHMMM BBBBB

4 0

3 years ago

Can someone please help

hram777 [196]

Answer: usuhduhduhdjhd

Explanation:

jaajaajaj

8 0

3 years ago