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3 years ago

Why isn't carbon-14 dating accurate for estimating the age of materials older than 50,000 years?

1 answer:

5 0

Carbon-14 is radioactive isotope of carbon.
Carbon is essential element of living cells. While the living cells are alive, the carbon contained in them are in equilibrium with the carbon in atmosphere. But, once the cell dies, the carbon-14 isotope undergoes radioactive decay. By measuring the carbon-14 in atmosphere to the carbon-14 in dead organism, we can calculate the time (or years) that organism have died.

However, carbon-14 dating technique is not accurate for estimating the age of materials older than 50,000 years old (above 40,000 years). This is because, 99% of carbon is carbon-12, 1% is carbon-13 and trace remaining is the carbon-14. This means, carbon-14 is found in very trace amount, in fact 1 part per trillion of carbon atoms present is carbon-14. The half of life of carbon-14 is 5,730 years. For dating the organism, we use the concept of half lives of the carbon-14 isotope in the dead organisms and calculate how many half life old the sample is. But as the years increases, the number of carbon-14 isotope becomes too low to detect and make accurate calculation.
This means, at some point the organism can simply run out of carbon-14.

Hence carbon-14 dating is not accurate for estimating age of materials older than 50,000 years old.

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According to the law of energy conservation, the heat lost is equal to the heat gained:

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$\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i$

We obtain:

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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.

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Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> smallest whole number ratio of elements in a compound

Molecular Formula => actual whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4' to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

% => grams => moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H => 7.14 : 14.29

To reduce mole values to the smallest whole number ratio, divide all mole values by the smaller mole value of the set.

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

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3 years ago