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madreJ [45]
3 years ago
13

Predict the chemical formula of the compound that will form when Li and o interact.

Chemistry
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

Li_2O

Explanation:

Hello!

In this case, when looking over the chemical formula of a compound formed when two elements interact, we need to realize about their location in the periodic table; thus, since lithium is a metal and oxygen a nonmetal, we infer lithium turns out the cation and oxygen the anion. Moreover, as the oxidation state of lithium is 1+ and that of oxygen is 2-, we can set up the chemical formula as follows:

Li^+O^{2-}\\\\Li_2O

Best regards!

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1s to the second power, 2s to the second power, 2p to the 6th power, 3s to the second power and 3p to the third power.
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I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w
trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

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5 0
3 years ago
A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of
vlada-n [284]

Answer: The molecular of the compound is, C_2H_3O

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=3.095g

Mass of H_2O=1.902g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.095g of carbon dioxide, \frac{12}{44}\times 3.095=0.844g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

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{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

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 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

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