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garri49 [273]
2 years ago
14

0.6 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24g of carbon, 0.04g of hydroge

n with the rest being oxygen. If the molecular weight of the substance is 90g/mol what is the emperical and molecular formula
Chemistry
1 answer:
prohojiy [21]2 years ago
5 0

Answer:

Molecular formula is, C₃H₆O₃

Empirical formula (with the lowest subscript) → CH₂O

Explanation:

We assume that the organic acid's mass is 0.6 g. We know that, in 0.6 g of compound we have 0.24 g of C and 0.04 g of H, then, we have

(0.6 - 0.24 - 0.04) = 0.32 g of O.

We determine each mol:

0.24 g /12 g/mol = 0.02 mol of C

0.04 g/ 1g/mol = 0.04 mol of H

0.32 g/ 16 g/mol = 0.02 mol of O

1 mol of acid weighs 90 g/mol.

In 0.6 g of acid, we have, (0.6 g / 90g/mol) = 0.0067 moles

Let's find out the formula with rules of three:

0.00667 mol of acid have 0.02 moles of C

1 mol of acid may have (0.02 /0.00667) = 3 mol of C

0.00667 mol of acid have 0.04 moles of H

1 mol of acid may have (0.04 /0.00667) =  6 mol of H

0.00667 mol of acid have 0.02 moles of O

1 mol of acid may have (0.02 /0.00667) = 3 mol of O

Molecular formula is, C₃H₆O₃

We confirm by the molar mass: 12g/mol . 3 + 6 . 1g/mol + 32g/mol . 3 = 90

Empirical formula (with the lowest subscript) → CH₂O

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The type of nuclear decay in which Oxygen decays to form nitrogen is beta plus decay.

<h3>What is a beta plus decay?</h3>

A beta plus decay is a type of decay in which a proton in an element disintegrates to produce a neutron resulting in a decrease in the atomic number of the radioactive element.

In the given equation below:

  • ^{15}_{7}O \rightarrow ^{15}_{7}N + ^{0}_{1}e

Oxygen decays to form nitrogen due to a decrease in atomic number.

This is an example of beta plus decay.

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<h3>Answer:</h3>

16.7 g H₂O

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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<u>Step 1: Define</u>

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

[Given] 1.85 mol NaOH

<u>Step 2: Identify Conversions</u>

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<u>Step 3: Stoichiometry</u>

  1. 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})
  2. Multiply/Divide:                 \displaystyle 16.6685 \ g \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

16.6685 g H₂O ≈ 16.7 g H₂O

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