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4 years ago

* Need ASAP * Explain the relationship between percent composition, Empirical formula, and molecular formula.

1 answer:

5 0

By dividing the percentage composition with the molar mass of that element we will get the empirical formula. Then using that empirical formula and formula mass we can find the molecular formula.

Explanation:

The chemical properties of any substance are defined obviously by the different types and relative amounts of atoms constituting its primary entities (in case of covalent compounds the primary entities are molecules and ions in the event of ionic compounds).

A percent composition of any compound gives the mass percent of each element present in the compound; in addition to that frequently it is determined experimentally and utilized to derive an empirical formula of any compound. An empirical formula mass of any covalent compound could be comparable with the molar or molecular mass of a compound to acquire a molecular formula.

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3 years ago

2C2H2 + 5O2 → 4CO2 + 2 H2O How many grams of CO2 are required to react with exactly 3.00 mol of O2? 106 g 132 g 165 g 76.8 g

cricket20 [7]

Hey there!:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

5 moles O2 ------------- 4 moles CO2

3.00 moles O2 ---------- ( moles of CO2 ?? )

moles of CO2 = 3.00 * 4 / 5

moles of CO2 = 12 / 5

moles of CO2 = 2.4 moles

So, molar mass CO2 = 44.01 g/mol

Therefore:

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mass of CO2 = 2.4 * 44.01 / 1

mass of CO2 = 106 g

Answer A

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3 years ago

A mixture of 10.0 g of Ne and 10.0 g Ar have a total pressure of 1.6 atm. What is the partial pressure of Ne?

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4 years ago

Read 2 more answers

In a 71.4 g sample of nahco 3 3 , how many grams of sodium are present?

sweet [91]

There are 19.5 g Na in 71.4 g NaHCO₃

Calculate the molecular mass of NaHCO₃.

1 Na = 1 × 22.99 u = 22.99 u

1 H = 1 × 1.008 u = 1.008 u

1 C = 1 × 12.01 u = 12.01 u

3 O = 3 × 16.00 u = 48.00 u

TOTAL = 84.008 u

So, there are 22.99 g of Na in 84.008 g NaHCO₃.

∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na

4 0

3 years ago