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4 years ago

What is the molecular formula of a compound with an empirical formula of C4H9 and a gram-formula mass of 114.0 g/mole?

1 answer:

6 0

The formula is easy:

Gram formula mass
----------------------------
Emprical Formula

So if we plug in our own numbers,

114.0
--------
57.116

We get an answer of 1.99, which rounds to 2.
Then, we distribute based off of our empirical formula.

2(C4H9) becomes C8H18.

Our molecular formula is C8H18.
Hope I could help!

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Why would scientists use a spectrometer in space.

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Answer:

A. To study the chemical composition of the surface of a planet

Explanation:

Strictly speaking, a spectrometer is any instrument used to view and analyze a range (or a spectrum) of a given characteristic for a substance (for example, a range of mass-to-charge values as in mass spectrometry), or a range of wavelengths as in absorption spectrometry like nuclear magnetic radiation spectroscopy

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2 years ago

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3 years ago

PLS HELPPP<br> why is CH3OH more polar than CH3CH2OH

dusya [7]

Answer:

Explanation:

Polarity is about differencens in electronegativity. CH bonds have around the same electronegativity value so a CH bond is nonpolar. The more CH bonds there are in a molecule, the more nonpolar it is. Since CH3CH2OH has more carbon-hydrogen bonds than CH3OH, it is more nonpolar. With the same reasoning, since CH3OH has less CH bonds, it's more polar.

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3 years ago

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M = 163 m/z) is formed.

DanielleElmas [232]

Answer:

As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.

Rule of Thirteen:

First divide the parent peak value by 13 as,

= 163 ÷ 13

= 12.53

Now, multiply 13 by 12,

= 13 × 12 (here, 12 specifies number of carbon atoms)

= 156

Now subtract 156 from 163,

= 163 - 156

= 7

Add 7 into 12,

= 7 + 12

= 19 (hydrogen atoms)

So, the rough formula we have is,

C₁₂H₁₉

Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,

C₁₂H₁₉ -------N--------> C₁₁H₁₇N

Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.

C₁₁H₁₇N -------O--------> C₁₀H₁₃NO

Molecular Formula: C₁₀H₁₃NO

Also,

In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.

The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.

It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.

The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.

6 0

4 years ago