a sample of copper wire with a mass of 1.659 g is heated with excess powdered sulfur until the copper has combined with the sulf
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1)Identify the atoms that are participating in a covalent bond.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.
Answer: The empirical formula is .
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of = 12.24 g
Mass of = 2.505 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 12.24 g of carbon dioxide, = of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 2.505 g of water, = of hydrogen will be contained.
Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H=
Moles of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O =
The ratio of C : H : O = 3: 3: 1
Hence the empirical formula is .