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3 years ago

The empirical formula for a compound is c2h5 and its formula mass is 58 amu. what is its molecular formula?

Chemistry

2 answers:

Nat2105 [25]3 years ago

4 0

Molecular formula of compound is C4H10

jasenka [17]3 years ago

4 0

Answer: $C_4H_{10}$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

Given empirical formula is $C_2H_5$ Empirical mass = $2\times 12+1\times 5=29amu$

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}=\frac{58}{29}=2$

The molecular formula will be $C_2H_5\times 2=C_4H_{10}$

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As you move across the period on the periodic table what pattern can be described about the electron configurations?

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Answer:

Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.

Explanation:

Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

Electronic configuration of lithium:

Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell and number of electron increased by 1.

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What are alleles A the decoders of the dna message B two forms of single genes c the basic unit inheritance d a measurable

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I think the answer is D

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3 years ago

Read 2 more answers

A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m

Lana71 [14]

Answer:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol$

Moles of hydrogen gas produced = 0.01225 mol

$2M+2xHCl\rightarrow 2MCl_x+xH_2$

Moles of metal = $\frac{0.225 g}{27.0 g/mol}=8.3333 mol$

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

$\frac{8.3333}{0.01225 mol}=\frac{2}{x}$

x = 2.9 ≈ 3

$2M+6HCl\rightarrow 2MCl_3+3H_2$

$MCl_3\rightarrow M^{3+}+Cl^-$

Formulas for the oxide and sulfate of M will be:

Oxide of M is $M_2O_3$ and sulfate of $M_2(SO_4)_3$ .

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Answer:

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3 years ago

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Answer:

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