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Nata [24]
3 years ago
7

Isoflurane, shown here, is a common inhalation anesthetic. write its molecular formula and calculate its percent composition by

mass
Chemistry
1 answer:
Natalka [10]3 years ago
7 0

Molecular formula of Isoflurane is <span><span>C</span></span>₃H₂ClF₅O.

Now calculate the percent composition by mass, which means percent of each element in the compound.

Mass of Isoflurane = 184.49 g/mol

Mass of carbon in the compound = 3 x 12.011 = 36.033g

Mass of hydrogen in the compound = 2 x 1.008 = 2.016g

Mass of chlorine in the compound = 1 x 35.453 = 35.453 g

Mass of fluorine in the compound = 5 x 18.998 = 94.99g

Mass of Oxygen in the compound = 1 x 16 = 16 g

Carbon’s percentage = Mass of carbon in the compound /mass of isoflurane x 100 =36.033/184.49 x 100 =19.53%

Hydrogen’s Percentage = Mass of hydrogen in the compound/mass of isoflurane x 100 = 2.016/184.49 = 1.09%

Chlorine’s percentage = Mass of chlorine in the compound/mass of isoflurane x 100 = 35.453/184.49 =19.22%

Flourine’s percentage = Mass of fluorine in the compound/mass of isoflurane x 100 = 94.99/184.49 x 100 = 51.49%

Oxygen’s percentage = Mass of Oxygen in the compound/mass of isoflurane x 100 =16/184.49 x 100 = 8.67%

<span> </span>

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What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

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

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

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

24.9235 L Ar ≈ 24.9 L Ar

5 0
3 years ago
Can someone help quick
alina1380 [7]

Answer:

Sodium

(Na)

Just count the electrons and search which atom it is.

7 0
2 years ago
A sample of 0.3283 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of A
Pavel [41]

Answer:

92.49 %

Explanation:

We first calculate the number of moles n of AgBr in 0.7127 g

n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g

n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol

Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and

From n = m/M

m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol

m = 0.0038 mol × 79.904 g/mol = 0.3036 g

% Br in compound = m₁/m₂ × 100%

m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)

m₂ = mass of compound = 0.3283 g

% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %

4 0
3 years ago
Two liquids are analyzed and found to both be 85.7% carbon and 14.3% hydrogen. At 750 mmHg and 150 C, both are gases. At these c
vodomira [7]

Answer:

Molecular formula A: C₅H₁₀

Molecular formula B: C₇H₁₄

Explanation:

It is possible to obtain empirical formula of compounds using percent composition, thus:

C: 85.7% × (1mol / 12.01g) = 7.136 moles C

H: 14.3% × (1mol / 1.01g) = 14.158 moles H

Mole ratio of H:C is:

14.158mol / 7.136mol = 2

That means in compounds A and B you have 2 hydrogens per atom of carbon and empirical formila is:

CH₂

Using PV = nRT, moles of A and B are:

<em>Where P is pressure (750mmHg / 760 = 0.987atm), V is volume (0.8000L), R is gas constant (0.082atmL/molK), and T is temperature (150°C +273.15 = 423.15K)</em>

Moles A and B: n = PV / RT

n = 0.987atm×0.8000L / 0.082atmL/molK×423.15K

n = 0.0228 moles of A and B.

Using the mass of A and B it is possible to find molar mass of each compound:

A = 1.60g / 0.0228mol = 70.31g/mol

B = 2.22g / 0.0228mol = 97.56g/mol

As empirical formula of both compounds is CH₂, (molar mass = 14.03g/mol). Molecular formula of compounds is:

A = 70.31g/mol / 14.03g/mol = 5 → Molecular formula: 5×CH₂ = <em>C₅H₁₀</em>

B = 97.56g/mol / 14.03g/mol = 7 → Molecular formula: 7×CH₂ = <em>C₇H₁₄</em>

8 0
3 years ago
What does the arrow in chemical equations mean? (PLEASE HELP)
grin007 [14]

Answer:

yields

Explanation:

I just did that quiz on k12 8th grade

6 0
3 years ago
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